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If the number 476 ** 0 is divisible by both 3 and 11, then in the hundredth and tenth places, the non-zero digits are,
respectively:
2 and 3
3 and 2
5 and 8
8 and 5
Let’s break this down step by step.
- We’re told 476 0 (let’s assume that’s a typo and they mean 476 _ * _0—"0" is a digit to find) is divisible by both 3 and 11.
- Divisibility by 3: sum of the digits must be a multiple of 3.
- Divisibility by 11: alternating sum (subtract one digit, add the next, etc.) must be a multiple of 11.
- The hundredth and tenth places in 476_0 are _ and 0.
- The options are giving you pairs of digits for those places. For each, we need the digits you put in to make it work.
- Let’s try each option:
Option 1: 2 and 3
47623
Sum: 4+7+6+2+3=22 (not divisible by 3)
Option 2: 3 and 2
47632
Sum: 22 (again, not divisible by 3)
Option 3: 5 and 8
47658
Sum: 4+7+6+5+8=30 (divisible by 3);
Alternating sum for 11: 4-7+6-5+8 = 6 (not divisible by 11)
Option 4: 8 and 5
47685
Sum: 4+7+6+8+5=30 (divisible by 3)
Alternating sum: 4-7+6-8+5 = 0 (divisible by 11)
The right answer is Option 4: 8 and 5.
What this really means is, the only pair that works for both divisibility rules is 8 (hundreds), 5 (tens).
The other options:
- Either don’t get you a sum divisible by 3, or
- Fail the alternating 11 rule.
- Option 4 is the only pair that actually fits the conditions.
By: santosh ProfileResourcesReport error
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