Difficulty: Easy
Correct Answer: Divisible by both 9 and 11
Explanation:
Introduction / Context:
This question asks you to apply quick divisibility tests to a large integer (58129745812974) without long division. The classic tests for 9 and 11 rely on digit sums and alternating sums, while the test for 4 depends on the last two digits only. Mastering these quick checks saves time on exams and interviews.
Given Data / Assumptions:
Concept / Approach:
Use well-known rules: a number is divisible by 9 if the sum of its digits is a multiple of 9; divisible by 11 if the alternating sum (sum of digits in odd positions minus sum of digits in even positions) is a multiple of 11 (including 0); divisible by 4 if the last two digits form a number divisible by 4.
Step-by-Step Solution:
1) Sum of digits for 9: 5+8+1+2+9+7+4+5+8+1+2+9+7+4 = 72. Since 72/9 = 8, the number is divisible by 9.2) Alternating sum for 11 (from leftmost as position 1): (5−8+1−2+9−7+4−5+8−1+2−9+7−4) = 0. A result of 0 means divisibility by 11.3) Test for 4: last two digits are 74; 74/4 is not an integer, so it is not divisible by 4.
Verification / Alternative check:
Because tests for 9 and 11 are independent, being divisible by both is consistent. A quick calculator check would confirm divisibility by 99 (since 9×11 = 99), but that is not necessary once the tests pass.
Why Other Options Are Wrong:
“Divisible by 9 only” and “Divisible by 11 only” ignore that both tests pass; “Divisible by 4” fails because 74 is not a multiple of 4; “Not divisible by any of these” is contradicted by both successful tests.
Common Pitfalls:
Forgetting to alternate signs correctly for 11; mis-adding a long digit sum; using only the last digit for 4 (the rule requires the last two digits).
Final Answer:
Divisible by both 9 and 11
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