JEE Mathematics Number Systems Notes Set 02

Divisibility Tests

Remainder rule: a number leaves the same remainder as its digit sum when divided by 3 or 9. Alternating-sum rule: counting places from the right, a number leaves the same remainder mod 11 as (sum of digits in odd places) minus (sum of digits in even places).

Worked Examples

1. Which of 2, 3, 4, 5, 6, 9, 11 divide 225324? It is even, digit sum 18, last two digits 24, and odd-place sum \(4+3+2=9\) equals even-place sum \(2+5+2=9\). So 2, 3, 4, 6, 9 and 11 divide it, but not 5.
2. Remainder of 3407 on division by 9: digit sum 14, and 14 leaves remainder 5, so the remainder is 5. For 218, digit sum 11 gives remainder 2. For 738, digit sum 18 gives remainder 0.
3. Find the digit x so that 7x563 is divisible by 11: \((7+5+3)-(x+6)=9-x=0\), so \(x=9\) (79563 = 11 times 7233).
4. Find x so that 8276x57 is divisible by 11: odd places from the right hold 7, x, 7, 8 and even places hold 5, 6, 2. The difference is \((22+x)-13=9+x\), which must be a multiple of 11, so \(x=2\).
5. Smallest number to add to 803642 to make it divisible by 9: digit sum 23, next multiple of 9 is 27, so add 4.
6. Smallest number to add to 803642 to make it divisible by 11: odd-place sum is \(2+6+0=8\), even-place sum is \(4+3+8=15\). Adding 7 to the units digit makes both 15 (803649), so add 7.