A prime number is a whole number greater than 1 whose only factors are 1 and itself.
Factors of 5 are 1, 5
Factors of 11 are 1, 11
Factors of 21 are 1, 3, 7, 21
Factors of 37 are 1, 37.
Hence, according to the definition of a prime number, 21 is not a prime number as it has more than two factors.
1 to 300 = 300 numbers
squares btwn 1 to 300 = 1,4,9,16,25,36,49.....289 i.e, total 17
cubes = 1,8,27,64,125,216 i.e, 5
now 1 and 64 are common so total numbers that should be removed ( 17+5-2= 20 )
total 20 term removed so 300th term would be 320
Sum of decimal places = 7.
Since the last digit to the extreme right will be zero (since 4 x 5 = 20), so there will be 6 significant digits to the right of the decimal point.
Screws weight = 90x100 = 9000gms
bolts weight = 100 x 150 = 15000gms
weight = screws + bolts => 24000 gms =>24 kg
Given entire box weight = 35.5kg
empty box = entire box weight - weight => 35.5kg - 24kg => 11.5kg
so the empty box is 11.5kg.
Last digit for the power of 6 is 6 (always)
Power cycle of 7 is 7, 9, 3, 1.
Now 467/4 gives a remainder of 3
Then the last digit is = 3
Last digit is 6 + 3 = 9.
Given number is 987 = 3 x 7 x 47.
So, required number must be divisible by each one of 3, 7, 47.
None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.
In case B, odd times even will give even
A perfect cube will have prime factors that are in groups of 3; for example 125 has the prime factors 5 x 5 x 5 , and 64 x 125 will also be a cube because its factors will be 4 x 4 x 4 x 5 x 5 x 5
Consider the answer choices in turn.
8 is the cube of 2, and p is a cube, and so the product will also be a cube.
pq will also be a cube as shown above.
pq is a cube and so is 27, but their sum need not be a cube. Consider the case where p =1 and q = 8, the sum of pq and 27 will be 35 which has factors 5 x 7 and is not a cube.
-p will be a cube.
Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).
1 1
8 A 9
+ 4 C 6
- 6 B 2
7 2 3
We may represent the given sum, as shown below:
1 + A + C - B = 12
A + C - B = 11
Now,giving the maximum values to A and C, i.e.
A = 9 and C = 9, we get B = 7.
Given
Thus, there are (5 + 5 + 9 + 9 + 3)= 31 prime numbers.
Let
x = peacocks
y = deers
2x + 4y = 200
x + y = 72
y = 72 - x
2x + 288 - 4x = 200
x = 44 peacocks
y = 28 deers
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