Question 1

Using each of the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once, how can you arrange them as a numerator and a denominator to form a fraction that is equal to 1/3?

Accepted Answer

Introduction / Context: This puzzle combines number sense, fractions and logical reasoning. You are asked to use each of the digits from 1 to 9 exactly once to form a numerator and a denominator such that the resulting fraction is equal to one third. Problems like this train you to think systematically about digits, divisibility and equivalent fractions, which is very helpful for both competitive exams and brain-teaser style interviews. Given Data / Assumptions: Available digits are 1, 2, 3, 4, 5, 6, 7, 8 and 9. Each digit must be used exactly once in the numerator and denominator together. The resulting fraction must be exactly equal to 1/3. Leading zeros are not allowed and the fraction should be written in standard numerator/denominator form. Concept / Approach: If a fraction N/D is equal to 1/3, then the denominator must satisfy D = 3 * N. Therefore, our task is to find a numerator N made from some of the digits 1 to 9 and a denominator D made from the remaining digits so that D is exactly three times N and all digits from 1 to 9 are used once. This is essentially a search problem with a strong arithmetic constraint. We can reduce the search space by using divisibility rules and length considerations for the two numbers. Step-by-Step Solution: Step 1: Let the fraction be N/D, with N using some digits from 1 to 9 and D using the remaining digits. Step 2: Because N/D = 1/3, we must have D = 3 * N. Step 3: We know that D must have more digits than N, because multiplying a positive integer by 3 usually increases its magnitude. A natural guess is that N is a four digit number and D is a five digit number, which together use all nine digits. Step 4: Systematically test possible four digit numerators made from 1 to 9, and for each N compute D = 3 * N. Check whether the resulting D uses exactly the remaining digits without repetition. Step 5: One such successful pair is N = 5823 and D = 17469. If you multiply 5823 by 3, you get 17469, and together the digits 5, 8, 2, 3, 1, 7, 4, 6, 9 are exactly the complete set from 1 to 9, each used once. Step 6: Therefore, the required fraction that satisfies all the conditions is 5823/17469, which equals 1/3. Verification / Alternative check: To verify, compute 5823 * 3. The product is 17469, which matches the denominator exactly, confirming that 5823/17469 = 1/3. You can also check that there is no repetition of digits: the numerator uses 5, 8, 2, 3 and the denominator uses 1, 7, 4, 6, 9. Combining them produces all digits 1 to 9 with no omissions or repeats. None of the other options, such as 17469/5823, give a value of one third. They yield fractions greater than 1 or with completely different values. Why Other Options Are Wrong: Option 17469/5823 is the reciprocal and equals 3, not 1/3. Option 2835/16947 does not satisfy the relation D = 3 * N, so its numerical value is not 1/3. Option 2583/17469 also fails the exact factor of 3 requirement and does not use digit placement correctly. Therefore these distractors do not meet the condition of forming a fraction exactly equal to one third. Common Pitfalls: Many learners try to guess by writing simple fractions like 123/369 and do not notice that digits repeat or that not all digits from 1 to 9 are used. Another common error is to check approximate decimal values instead of enforcing the exact relationship D = 3 * N. Testing random combinations without a strategy is time consuming and can lead to frustration. A more systematic approach, based on the algebraic condition and careful tracking of digits, is far more effective for such puzzles. Final Answer: The fraction using each digit 1 to 9 exactly once that equals one third is 5823/17469.

Question 2

Which mathematical symbol can be placed between 5 and 9 to form a single number that is greater than 5 but smaller than 9?

Accepted Answer

Introduction / Context: This classic puzzle checks your understanding of numbers, symbols and place value rather than simple arithmetic. You are asked to place a mathematical symbol between 5 and 9 such that the result is a number greater than 5 but less than 9. At first glance, it might appear that any operation combining 5 and 9 will either make the value much bigger than 9 or smaller than 5, but the trick lies in thinking beyond basic operations like addition and subtraction. Given Data / Assumptions: We start with the digits 5 and 9 written side by side. We are allowed to insert a mathematical symbol between them. The final result must be a single number whose value is greater than 5 and smaller than 9. The options suggest several possible symbols, such as a decimal point, plus sign, division sign and comparison sign. Concept / Approach: The key idea is to remember that mathematical symbols are not limited to arithmetic operations like plus or minus. A decimal point is also a mathematical symbol, and when placed between two digits, it forms a decimal number. By placing a decimal point between 5 and 9, we get 5.9, which clearly lies between 5 and 9 on the number line. In contrast, placing arithmetic or comparison operators between the digits usually produces an expression, not a single number in the required interval. Step-by-Step Solution: Step 1: Consider placing a plus sign: 5+9 gives 14, which is greater than 9 and therefore does not satisfy the condition. Step 2: Consider placing a division sign: 5÷9 gives a decimal less than 1, which is smaller than 5 and so this option fails. Step 3: Consider placing a greater-than sign: 5>9 is not a single number at all, it is just a false statement, and therefore does not match the puzzle requirement. Step 4: Now consider placing a decimal point between the digits to form 5.9. This is a single decimal number lying strictly between 5 and 9. Step 5: Check the condition: 5.9 is greater than 5 and smaller than 9, so this choice satisfies the puzzle exactly. Verification / Alternative check: To verify, imagine a number line from 0 to 10. The number 5 is to the left and 9 is to the right. The decimal 5.9 is between 5 and 6, so it clearly lies between 5 and 9. None of the other options produce a single number in this interval. In exam situations, checking quickly on a mental number line is a reliable way to confirm that your answer fits the inequality conditions. Why Other Options Are Wrong: The plus sign produces 14, which is far greater than 9. The division sign produces a result less than 1, so it does not lie between 5 and 9. The greater-than symbol produces a comparison statement, not a numeric value, and is also logically incorrect as 5 is not greater than 9. Therefore, only the decimal point meets the specific requirement of forming one valid number in the correct range. Common Pitfalls: Many learners initially focus only on arithmetic operators and try combinations such as 9−5, 9÷5 or 5+9, forgetting that the puzzle explicitly talks about placing a symbol between 5 and 9, not necessarily performing an operation in the usual way. Another pitfall is to interpret the question as requiring an equation instead of a single number. The best way to avoid these mistakes is to think flexibly about what counts as a mathematical symbol and to test the resulting value against the inequality condition. Final Answer: The correct symbol is a decimal point, forming the number 5.9, which is greater than 5 and smaller than 9.

Question 3

In the number pattern 111 13, 112 24, 113 35, 117 ?, what number should replace the question mark in the final pair?

Accepted Answer

Introduction / Context: This question is a classic number sequence and pattern recognition puzzle. You are given pairs of numbers, such as 111 13, 112 24 and 113 35, and asked to determine the missing number that should be paired with 117. Such puzzles commonly appear in reasoning sections of competitive exams and are designed to test your ability to detect hidden relationships between numbers, including digit sums and positional patterns. Given Data / Assumptions: We are given the pairs (111, 13), (112, 24), (113, 35) and (117, ?). The question mark stands for an unknown two digit number to be determined. The relationship between each three digit number and its corresponding two digit number is consistent and must be discovered. The options provided are 46, 57, 68 and 79. Concept / Approach: In many such puzzles, the smaller number in each pair is related to the digits of the larger number through operations like digit sum, differences or rearrangement. Here, observing the pairs carefully suggests that the two digit number can be thought of as two separate digits with specific meanings. In particular, the first digit of the two digit number matches the last digit of the three digit number, and the second digit is equal to the sum of the digits of the three digit number. Recognising this simple but non obvious rule is the key to solving the sequence. Step-by-Step Solution: Step 1: Look at the first pair 111 and 13. The digits of 111 are 1, 1 and 1. The sum of these digits is 1+1+1 = 3. The last digit of 111 is 1. Putting the last digit and the sum together gives 1 and 3, which form 13. Step 2: Examine the second pair 112 and 24. The digits of 112 are 1, 1 and 2. Their sum is 1+1+2 = 4. The last digit of 112 is 2. Putting the last digit and the sum together gives 2 and 4, which form 24. Step 3: Consider the third pair 113 and 35. The digits of 113 are 1, 1 and 3. Their sum is 1+1+3 = 5. The last digit of 113 is 3. Combining last digit 3 and sum 5 gives 35, matching the given pair. Step 4: Now apply the same rule to 117. The digits are 1, 1 and 7. Their sum is 1+1+7 = 9. The last digit of 117 is 7. Joining last digit 7 and sum 9 gives the number 79. Step 5: Check the options: 79 is present as one of the answer choices, so this value fits the discovered pattern perfectly. Verification / Alternative check: As a quick verification, apply the rule back to each of the earlier pairs. For 111, sum of digits is 3 and last digit is 1, giving 13. For 112, sum of digits is 4 and last digit is 2, giving 24. For 113, sum is 5 and last digit is 3, giving 35. Since the same pattern works for all given pairs, it is very likely that this is the intended rule. Applying it consistently to 117 produces 79, so the rule is stable and the answer is reliable. Why Other Options Are Wrong: The options 46, 57 and 68 do not follow the discovered logic. For 117, the last digit is 7, so the first digit of the correct two digit number must be 7. That alone rules out 46, 57 and 68, because none of them start with 7. Even if you try to force another rule, those options do not provide a consistent mapping for the earlier pairs, so they cannot represent the intended solution to the sequence. Common Pitfalls: Learners sometimes focus only on simple differences between the two digit numbers, for example noting that 13, 24 and 35 all increase by 11, but this does not meaningfully connect to the three digit numbers. Others may try random arithmetic operations like dividing the large number by a constant, which quickly becomes messy. The better strategy in sequence questions is to relate the digits of each number in a pair and to look for patterns involving sums or positions of digits. That leads quickly to the simple rule involving digit sum and last digit, which explains all the examples clearly. Final Answer: The number that should replace the question mark in the pair with 117 is 79.

Question 4

In this number puzzle, 1+1+1+1 gives FOUR → R, 2+2+2+2 gives EIGHT → T, 3+3+3+3 gives TWELVE → E, 4+4+4+4 gives SIXTEEN → N. Then 5+5+5+5 corresponds to which final letter?

Accepted Answer

Introduction / Context: This is a classic number and word pattern puzzle. At first glance, the expression "1 1 1 1 = R, 2 2 2 2 = T, 3 3 3 3 = E, 4 4 4 4 = N" looks mysterious, because it connects repeated digits with letters. The key to solving the puzzle is to realise that these repeated digits hint at addition and then at the English word for the resulting number. The letter on the right side comes from that word. The question then asks you to extend this pattern logically to "5 5 5 5" and decide which letter it should correspond to. Given Data / Assumptions: We are given four examples: 1 1 1 1 → R, 2 2 2 2 → T, 3 3 3 3 → E, 4 4 4 4 → N. Each group of four identical digits is intended to be combined with addition, for example 1+1+1+1. The sum of the four identical digits is converted into an English number word (such as FOUR or EIGHT). The displayed capital letter (R, T, E, N) is taken from this English word in a consistent way. We must find the corresponding letter for the case 5 5 5 5 using the same hidden rule. Concept / Approach: The central idea is to translate the repeated digits into arithmetic, then into words, and finally into letters. First, treat "1 1 1 1" as 1+1+1+1, which equals 4. The number 4 is written as FOUR in English, and the given letter is R, the last letter of FOUR. Next, 2+2+2+2 equals 8. The number 8 is written as EIGHT, and the last letter of EIGHT is T, which matches the puzzle. Similarly, 3+3+3+3 is 12, written as TWELVE, whose last letter is E; and 4+4+4+4 is 16, written as SIXTEEN, whose last letter is N. In each case the letter on the right is the last letter of the English word for the sum. Step-by-Step Solution: Step 1: Convert the pattern "1 1 1 1" into an addition: 1+1+1+1 = 4 → FOUR → last letter R. This.matches the given mapping. Step 2: Convert "2 2 2 2" into 2+2+2+2 = 8 → EIGHT → last letter T. This also matches the given mapping. Step 3: Convert "3 3 3 3" into 3+3+3+3 = 12 → TWELVE → last letter E, which again matches the puzzle. Step 4: Convert "4 4 4 4" into 4+4+4+4 = 16 → SIXTEEN → last letter N, matching the final given letter. Step 5: Apply the same rule to "5 5 5 5". Compute 5+5+5+5 = 20. Step 6: Write 20 in English as TWENTY. The last letter of TWENTY is Y. Step 7: Therefore, according to exactly the same pattern, 5 5 5 5 must correspond to the letter Y. Verification / Alternative check: To verify the rule, you can recheck every given example and confirm that no other consistent interpretation works as cleanly. In all four base cases, the process of summing the four digits, converting to the English word, and then taking the final letter gives exactly the listed letters R, T, E, and N. Furthermore, there is no need to change the rule halfway through, since 5+5+5+5 is a natural continuation of the sequence. When we compute the sum as 20 and write TWENTY, the last letter is clearly Y, which fits neatly with the pattern seen before and does not contradict any of the given mappings. Why Other Options Are Wrong: Option Z does not correspond to any obvious feature of the word TWENTY. Option T might tempt some learners because T appears in TWENTY, but the rule is clearly about the last letter, not just any letter. Option X has no connection to any of the words FOUR, EIGHT, TWELVE, SIXTEEN, or TWENTY. Thus these alternatives do not follow the established pattern. Only Y arises naturally from the sustained logic of the puzzle and agrees with all earlier examples. Common Pitfalls: A common mistake is to look for patterns directly in the digits 1, 2, 3, and 4, or to treat the letters R, T, E, and N as initial letters of unknown words. Another pitfall is to focus on multiplication or concatenation (for example, 1111, 2222) instead of seeing that the repetition hints at repeated addition. When approaching such puzzles, it is useful to test small guesses like adding the digits and then thinking of number names. This strategy often reveals hidden word based patterns like the one in this question. Final Answer: Following the pattern, 5+5+5+5 = 20 → TWENTY, whose last letter is Y.

Question 5

Find the unique 10-digit self-descriptive number in which the 1st digit counts zeros, the 2nd counts ones, … and the 10th digit counts nines in the number itself.

Accepted Answer

Introduction / Context: This puzzle asks you to find a very special 10-digit number with a self-descriptive property. Each position in the number tells you how many times a particular digit appears in the entire number. The first digit tells you how many zeros there are, the second digit tells you how many ones there are, the third digit tells you how many twos there are, and so on, until the tenth digit which tells you how many nines there are. Numbers like this are called self descriptive numbers and they are an excellent test of logical reasoning, counting, and consistency checking. Given Data / Assumptions: The number must be exactly 10 digits long. The 1st digit equals the total count of digit 0 in the entire 10-digit number. The 2nd digit equals the total count of digit 1 in the number. The 3rd digit equals the total count of digit 2, and so on. The 10th digit equals the total count of digit 9 in the number. All counts must be non negative integers, and the total of all counts must be 10, because the number has 10 digits in total. Concept / Approach: The concept is to build a self consistent system of counts. We are really solving a system of conditions: the digit in position i tells us how many of digit (i - 1) appear. For 10 digits, the counts of 0 through 9 must add up to 10. In known self descriptive numbers in base 10, most digits are zero, and only a few are non zero counts. A standard solution is 6210001000. In this candidate, the digits from left to right are 6, 2, 1, 0, 0, 0, 1, 0, 0, 0. We must verify that this 10-digit value satisfies the rules that define the puzzle. Step-by-Step Solution: Step 1: Write the candidate number clearly: 6210001000. Step 2: Count how many zeros appear in the entire number. The digits that are not zero are 6, 2, 1, and 1 (four digits). The total length is 10, so there are 10 - 4 = 6 zeros. The first digit is 6, so the count of zeros matches the first digit. Step 3: Count how many ones appear. Looking through 6210001000, there are exactly two digits 1. The second digit of the number is 2, which correctly records that there are two ones. Step 4: Count how many twos appear. There is exactly one digit 2 in the whole number. The third digit is 1, which correctly records one two. Step 5: Count how many threes appear. There are no 3 digits in 6210001000. The fourth digit is 0, so the count of threes is correctly recorded as zero. Step 6: Similarly, count how many fours appear. There are no 4 digits, and the fifth digit is 0, so this also matches. Step 7: Count how many fives appear. Again, there are no 5 digits, and the sixth digit is 0, which is consistent. Step 8: Count how many sixes appear. There is exactly one 6 in the entire number, at the first position. The seventh digit is 1, so the count of sixes is correctly recorded. Step 9: Count how many sevens, eights, and nines appear. None of these digits occur anywhere in 6210001000, so each of their counts is 0. The eighth, ninth, and tenth digits are 0, 0, and 0, which is fully consistent. Step 10: Check that the sum of all the counts equals 10: 6 (zeros) + 2 (ones) + 1 (twos) + 0 + 0 + 0 + 1 (sixes) + 0 + 0 + 0 = 10. This matches the total number of digits, further confirming that the number is valid. Verification / Alternative check: To verify more systematically, you can write a small table with digits 0 to 9 in one row and the corresponding counts implied by 6210001000 in the second row. For each digit, check the actual frequency in the number. You will find that digit 0 occurs six times, digit 1 occurs two times, digit 2 occurs once, digits 3, 4, and 5 occur zero times, digit 6 occurs once, and digits 7, 8, and 9 occur zero times. These observed counts exactly match the digits 6, 2, 1, 0, 0, 0, 1, 0, 0, 0 in the candidate number. Because all conditions are satisfied, 6210001000 is a self descriptive number and answers the puzzle completely. Why Other Options Are Wrong: In 5210011000, the first digit claims that there are five zeros, but a quick count reveals a different number of zeros, so it fails the very first condition. In 4210102000 and 6110002000, similar inconsistencies appear: for example the claimed numbers of ones or twos do not match their actual frequencies in the number. Any deviation between a claimed count and the real number of occurrences immediately disqualifies a candidate. Only 6210001000 survives all such checks. Common Pitfalls: A common mistake is to try to guess a pattern rather than check counts carefully. Many learners also forget that the total of all digits in the number must equal 10, because they are counting how often each digit from 0 to 9 appears. Another pitfall is to miscount zeros, especially when they appear in long blocks. A reliable method is to list all digits, circle non zero entries, and then calculate the number of zeros as the remainder. Being systematic and patient is essential in self descriptive number puzzles. Final Answer: The required 10-digit self descriptive number is 6210001000, where each position correctly counts how many times its corresponding digit appears in the entire number.

Question 6

Number puzzle: which 5-digit number becomes exactly three times larger when the digit 1 is written at the end instead of at the beginning?

Accepted Answer

Introduction / Context: This classic number puzzle tests logical reasoning with place value and simple algebra. We are given a special 5 digit number such that writing the digit 1 in front of it produces a number that is exactly three times smaller than the number obtained by writing the digit 1 at the end. Questions like this often appear in aptitude tests and bank or SSC style exams to check comfort with numbers and proportional relationships. Given Data / Assumptions: The original number is a 5 digit integer. Let us call it N. Writing 1 at the beginning gives a 6 digit number, written as 1N. Writing 1 at the end gives a 6 digit number, written as N1. The number with 1 at the end is three times the number with 1 at the beginning. We assume standard base 10 notation and no leading zeros in N. Concept / Approach: We convert the verbal description into equations using place value. If N is 5 digits, then putting 1 in front is 1 * 10^5 + N, and putting 1 at the end is 10 * N + 1. The condition that the second is three times the first becomes a linear equation in N. Solving this equation gives the required number. Finally, we verify that the solution is a 5 digit integer and that it indeed satisfies the given relationship exactly. Step-by-Step Solution: Let N be the unknown 5 digit number. Number with 1 at the front = 1 * 10^5 + N = 100000 + N. Number with 1 at the end = 10 * N + 1. Given condition: 10 * N + 1 = 3 * (100000 + N). Expand right side: 10 * N + 1 = 300000 + 3 * N. Rearrange: 10 * N - 3 * N = 300000 - 1. So 7 * N = 299999. Therefore N = 299999 / 7 = 42857. Check that N is 5 digits and satisfies the original wording. Verification / Alternative check: Compute the two 6 digit numbers explicitly. With N = 42857, putting 1 in front gives 142857. Putting 1 at the end gives 428571. Now check the ratio: 428571 / 142857 = 3 exactly, with no remainder. This shows that the number obtained by placing 1 at the end is exactly three times the number obtained by placing 1 at the beginning. Also, 42857 clearly has 5 digits, so it satisfies all required conditions. No other 5 digit integer will satisfy the linear equation 7 * N = 299999, so the solution is unique. Why Other Options Are Wrong: Option 14285 does not work: 114285 * 3 equals 342855, which is not equal to 142851. Option 28571 fails in the same way because the ratio between 128571 and 285711 is not exactly three. Option 42168 also does not satisfy the algebraic equation because it does not solve 7 * N = 299999. Only 42857 produces the precise three times relationship demanded in the problem statement. Common Pitfalls: A common error is to misinterpret three times smaller and set up the proportion incorrectly, for example by writing 1N = 3 * N1 instead of the correct 10 * N + 1 = 3 * (100000 + N). Another frequent mistake is mishandling powers of 10 and place values when constructing 1N and N1. Some students also attempt guess and check with random 5 digit numbers, which is inefficient and prone to arithmetic mistakes. Setting up the clear algebraic equation and solving systematically is the safest and fastest method in exam conditions. Final Answer: The required 5 digit number is 42857.

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