The 6-digit number 72k23l is divisible by 88. Find the digits k and l that make the number exactly divisible by both 8 and 11.

Introduction:
This question tests knowledge of divisibility rules, especially for composite numbers like 88. Since 88 = 8 * 11, a number is divisible by 88 if and only if it is divisible by both 8 and 11. You are given a 6-digit number with two unknown digits k and l and must determine these digits so that the number 72k23l is divisible by 88.

Given Data / Assumptions:

• The number is 72k23l, where k and l are single digits from 0 to 9.
• The number is divisible by 88.
• 88 = 8 * 11, so the number must be divisible by 8 and also by 11.
• We will use the divisibility rules for 8 and 11.

Concept / Approach:
First, use the rule for divisibility by 8: a number is divisible by 8 if its last three digits form a number divisible by 8. This gives a condition on l. Then apply the rule for 11: the difference between the sum of digits in odd positions and the sum of digits in even positions must be a multiple of 11 (including 0). This gives a condition on k once l is fixed. Solving both conditions together yields k and l.

Step-by-Step Solution:
Divisibility by 8: look at the last three digits, which form 23l.This 3-digit number is 230 + l and must be divisible by 8.Try l from 0 to 9. Check 232: 232 / 8 = 29, which is an integer.So l = 2 is the only digit making 23l divisible by 8.Now the number is 72k232.Divisibility by 11: consider digits 7, 2, k, 2, 3, 2 from left to right.Odd positions (1st, 3rd, 5th): 7, k, 3 => sum = 7 + k + 3 = k + 10.Even positions (2nd, 4th, 6th): 2, 2, 2 => sum = 2 + 2 + 2 = 6.For divisibility by 11, difference (odd sum - even sum) must be a multiple of 11.Difference = (k + 10) - 6 = k + 4.So k + 4 must be 0, 11, 22, etc. With k a digit, k + 4 = 11 is valid, giving k = 7.Thus k = 7 and l = 2.

Verification / Alternative check:
Check last three digits: 232 / 8 = 29, so divisible by 8. Now check divisibility by 11 using 727232: sum of odd-position digits = 7 + 7 + 3 = 17; sum of even-position digits = 2 + 2 + 2 = 6; difference = 11, which is a multiple of 11. Therefore 727232 is divisible by 11 and hence by 88.

Why Other Options Are Wrong:
k = 8 and l = 2; k = 8 and l = 3; k = 7 and l = 1; k = 6 and l = 4: in each of these combinations, either the last three digits are not divisible by 8 or the 11-divisibility rule fails, so the full number cannot be divisible by 88.

Common Pitfalls:
Trying to check divisibility by 88 directly without using the simpler separate tests for 8 and 11.Forgetting that only the last three digits matter for divisibility by 8.Miscomputing the alternating sums for the 11-divisibility rule by mixing up digit positions.

Final Answer:
k = 7 and l = 2