6. Divisibility Rules Applied to Find Missing Digits

This is a very frequent SSC/RRB question format: “Find the digit x so that the number ___ is divisible by ___.”

Method: Apply the relevant divisibility rule treating the unknown digit as a variable, then solve for the digit (0–9) that satisfies the condition.

Example 1 (divisible by 9): Find x if 27x4 is divisible by 9. Digit sum = 2+7+x+4 = 13+x. For divisibility by 9, 13+x must be a multiple of 9 → nearest multiple ≥13 is 18 → x=5. Check: 2754 ÷ 9 = 306 exactly. ✓

Example 2 (divisible by 11): Find x if 46x27 is divisible by 11. (Worked fully in Section 2.10, Example 3.) Answer: x=8.

Example 3 (two missing digits, divisible by 8 and 9, i.e., by 72): Find x if the number 1x2536 is divisible by 72. Since 72=8×9, check both: Divisible by 8: last three digits = 536. 536 ÷ 8 = 67 exactly → this condition is automatically satisfied regardless of x (since x doesn’t appear in the last 3 digits). Divisible by 9: digit sum =1+x+2+5+3+6=17+x. Nearest multiple of 9 ≥17 is 18 → x=1. So the number is 112536. Check: 112536 ÷ 72 = 1563 exactly. ✓

Example 4 (three-digit divisor test combined with digit sum): Find x if the number 5x652 is divisible by 4. Divisibility by 4 depends only on the last two digits, “52,” which is already divisible by 4 (52÷4=13) regardless of x. This means every digit x from 0 to 9 works — a useful reminder that not every missing-digit question has a unique answer; sometimes the rule doesn’t even involve the unknown digit, and the question is testing whether you know that.

Example 5 (genuine two-missing-digit problem, both blanks actually solved): Find the digits x and y so that the 6-digit number 5x473y is divisible by 72. Since 72=8×9, split the work: Divisible by 8: last three digits are “73y” = 730+y. We need 730+y ≡ 0 (mod 8). Since 730 = 8×91+2, this becomes 2+y≡08, i.e., y≡-2≡68. In the digit range 0–9, only y=6 works (since y=14 is not a valid digit). Divisible by 9: digit sum =5+x+4+7+3+y=19+x+y=19+x+6=25+x. Nearest multiple of 9 ≥25 is 27 → x=2. So x=2, y=6, giving the number 5,24,736. Check: 524736÷72=7288 exactly, and separately 524736÷8=65592, 524736÷9=58304, both exact. ✓ x+y=8.

Example 6 (edge case — a two-blank problem with more than one valid pair): Find the digits x and y (with y 0,5, since the number must be divisible by 5) so that 3x67y is divisible by 45. Since 45=9×5, the last digit y must be 0 or 5. - If y=0: digit sum =3+x+6+7+0=16+x. Nearest multiple of 9 ≥16 is 18 → x=2, giving 32,670. - If y=5: digit sum =3+x+6+7+5=21+x. Nearest multiple of 9 ≥21 is 27 → x=6, giving 36,675. Both 32670÷45=726 and 36675÷45=815 check out exactly. So two different pairs (x,y)=(2,0) and (6,5) both work — a good reminder that when a missing-digit condition (like “y = 0 or 5”) itself has two branches, you may need to solve the problem once per branch rather than assuming a single final answer.