Divisibility Rules and Tests
Divisibility Rules and Tests
Divisibility is the most-asked Number System pattern in RRB NTPC. Missing-digit questions for 11 and 72 have come in both stages: the 2022 CBT-2 papers and the August 2025 UG CBT-1. Each one is a 25-second question if the rule is automatic and a 2-minute question if you try long division.
1. Complete Divisibility Rule Table
| Divisor | Rule | Quick check |
|---|---|---|
| 2 | Last digit even (0, 2, 4, 6, 8) | 3,578 yes |
| 3 | Sum of digits divisible by 3 | 4,512: 4+5+1+2 = 12 yes |
| 4 | Last two digits divisible by 4 (or 00) | 7,316: 16 yes |
| 5 | Last digit 0 or 5 | 9,435 yes |
| 6 | Divisible by both 2 and 3 | 1,254: even, sum 12 yes |
| 7 | Double the last digit, subtract from the rest; repeat | 343: 34 − 6 = 28 yes |
| 8 | Last three digits divisible by 8 (or 000) | 51,128: 128 yes |
| 9 | Sum of digits divisible by 9 | 7,425: sum 18 yes |
| 10 | Last digit 0 | 4,560 yes |
| 11 | (Sum of digits at odd places) − (sum at even places) = 0 or a multiple of 11 | 9,163: (9+6) − (1+3) = 11 yes |
| 12 | Divisible by 3 and 4 | 4,236: sum 15, last two 36 yes |
| 13 | Add 4 × last digit to the rest; repeat | 247: 24 + 28 = 52 = 13 × 4 yes |
| 16 | Last four digits divisible by 16 | 3,20,048: 0048 yes |
| 18 | Even and divisible by 9 | 7,236 yes |
| 25 | Last two digits 00, 25, 50 or 75 | 6,175 yes |
| 36 | Divisible by 4 and 9 | 4,572 yes |
| 72 | Divisible by 8 and 9 | 5,32,728 yes |
| 99 | Sum of two-digit groups from the right divisible by 99 | 3,35,214: 33 + 52 + 14 = 99 yes |
| 125 | Last three digits divisible by 125 | 4,375 yes |
Rule for a composite divisor: split it into CO-PRIME factors and test each. 24 = 3 × 8, not 4 × 6 (4 and 6 share 2). A number divisible by 4 and 6 need not be divisible by 24 (example: 12).
✗ Divisible by 4 and 6, so divisible by 24 | ✓ Use co-prime pair 3 and 8; 36 is divisible by 4 and 6 but not 24
2. The 1001 Trick (7, 11, 13 together)
7 × 11 × 13 = 1001. Split the number into groups of three from the right and take the alternating sum of the groups. If that result is divisible by 7, 11 or 13, so is the whole number.
4,59,319 → 459 − 319 = 140 → 140 = 7 × 20, so the number is divisible by 7 (not by 11 or 13)
Any six-digit number of the form abcabc = abc × 1001, so it is always divisible by 7, 11 and 13. A three-digit number aaa = a × 111 = a × 3 × 37, so it is always divisible by 3 and 37.
3. Algebraic Divisibility Facts
| Expression | Always divisible by |
|---|---|
| aⁿ − bⁿ (any n) | a − b |
| aⁿ − bⁿ (n even) | a − b and a + b |
| aⁿ + bⁿ (n odd) | a + b |
| aⁿ + bⁿ (n even) | Neither a + b nor a − b in general |
| Product of n consecutive integers | n! (so product of 3 consecutive is divisible by 6) |
| n³ − n | 6 (it is (n − 1)n(n + 1)) |
| ab − ba (two-digit) | 9 |
| ab + ba (two-digit) | 11 |
Counting multiples: numbers from 1 to N divisible by k = ⌊N/k⌋ (integer part). Numbers in a range = ⌊upper/k⌋ − ⌊(lower − 1)/k⌋ when both ends are included.
हिंदी नोट: 11 से विभाज्यता के लिए विषम स्थानों के अंकों का योग और सम स्थानों के अंकों का योग निकालिए। दोनों का अंतर 0 या 11 का गुणज हो तो संख्या 11 से विभाज्य है। स्थान बाएँ से गिनें या दाएँ से, उत्तर वही आता है।
Exam Pointer: Divisibility is a regular NTPC question source. Verified patterns: missing digit for 11 or 9, which divisor divides a long number, counting multiples in a range, largest or smallest n-digit multiple, and the least number to add or subtract. The trap is a second condition (divisible by 72, or "least" vs "greatest" value) that makes the obvious digit wrong.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Missing digit for divisibility by 11 (one or two unknowns)
[PYQ: NTPC UG CBT-1 11-Aug-2025 Shift-3 | NTPC CBT-2 10-May-2022 Shift-1]
EXAM LEVEL
Q. The seven-digit number 583x426 is divisible by 11. Find the digit x.
Odd places from the left (1st, 3rd, 5th, 7th): 5 + 3 + 4 + 6 = 18. Even places (2nd, 4th, 6th): 8 + x + 2 = 10 + x. The difference 18 − (10 + x) = 8 − x must be 0 or ±11. Since x is a single digit, 8 − x = 0 and x = 8. Check: 5,838,426 = 11 × 530,766.
Answer: x = 8
EXAMATLAS LEVEL
Q. The six-digit number 3x52y4 is divisible by both 9 and 11. Find x × y.
For 9: 3 + x + 5 + 2 + y + 4 = 14 + x + y must be a multiple of 9, so x + y = 4 or 13. For 11: odd places 3 + 5 + y = 8 + y, even places x + 2 + 4 = 6 + x, difference 2 + y − x must be 0 or ±11. The ±11 cases need y − x = 9 (x = 0, y = 9, sum 9, rejected) or x − y = 13 (impossible). So x = y + 2. With x + y = 4 this gives y = 1, x = 3; with x + y = 13 it gives a fraction. The number is 335214, and 335214 = 99 × 3,386 confirms it. x × y = 3.
Answer: 3
Pattern 2: Least or greatest digit for divisibility by 3, 9, 6, 8 or 72
[PYQ: NTPC CBT-2 14-Jun-2022 Shift-2 | NTPC 2020-21 cycle]
EXAM LEVEL
Q. Find the least value of k for which 472k85 is divisible by 9.
Digit sum = 4 + 7 + 2 + k + 8 + 5 = 26 + k. The next multiple of 9 above 26 is 27, so k = 1.
Answer: 1
EXAMATLAS LEVEL
Q. The number 53p72q is divisible by 72. Find the greatest possible value of p + q.
72 = 8 × 9 (co-prime). For 8 the last three digits 72q must be divisible by 8; 720 = 8 × 90, so q = 0 or 8. For 9: 5 + 3 + p + 7 + 2 + q = 17 + p + q is a multiple of 9, so p + q = 1 or 10. With q = 0, p = 1 (sum 1). With q = 8, p = 2 (sum 10). Greatest p + q = 10, and 532728 = 72 × 7,399 confirms it. Students who stop at the first solution mark 1.
Answer: 10
Pattern 3: Which divisor divides a given large number
[PYQ: NTPC Graduate CBT-1 6-Jun-2025 Shift-2 | NTPC CBT-2 12-Jun-2022 Shift-1]
EXAM LEVEL
Q. Which of the following divides 4,59,319 exactly? (a) 3 (b) 7 (c) 9 (d) 11
Digit sum is 31, so 3 and 9 are out. For 11: (4 + 9 + 1) − (5 + 3 + 9) = −3, out. For 7 use the 1001 trick: 459 − 319 = 140 = 7 × 20, so it divides.
Answer: (b) 7
EXAMATLAS LEVEL
Q. Which of the following is NOT a factor of 8,61,861? (a) 49 (b) 123 (c) 143 (d) 187
861861 is of the form abcabc, so it equals 861 × 1001. Now 861 = 3 × 7 × 41 and 1001 = 7 × 11 × 13, giving 861861 = 3 × 7² × 11 × 13 × 41. 49 = 7² is present, 123 = 3 × 41 is present, 143 = 11 × 13 is present, but 187 = 11 × 17 needs a 17 that is absent.
Answer: (d) 187
Pattern 4: Counting multiples in a range
[PYQ: NTPC 2020-21 cycle]
EXAM LEVEL
Q. How many numbers between 200 and 900 are divisible by 7?
Multiples of 7 up to 899: ⌊899/7⌋ = 128. Multiples up to 200: ⌊200/7⌋ = 28. Required count = 128 − 28 = 100. Neither end point is a multiple of 7, so "between" causes no adjustment here.
Answer: 100
EXAMATLAS LEVEL
Q. How many numbers from 1 to 1,000 are divisible by 4 or 6 but not by both?
Divisible by 4: ⌊1000/4⌋ = 250. Divisible by 6: ⌊1000/6⌋ = 166. Divisible by both means divisible by LCM 12: ⌊1000/12⌋ = 83. "Either but not both" = 250 + 166 − 2 × 83 = 250. The usual slip subtracts 83 only once, giving 333, which is "4 or 6" including both.
Answer: 250
Pattern 5: Largest or smallest n-digit multiple
[PYQ: NTPC CBT-1 4-Jan-2021 Shift-2]
EXAM LEVEL
Q. Find the largest four-digit number exactly divisible by 47.
9,999 ÷ 47 gives quotient 212 and remainder 35 (47 × 212 = 9,964). Subtract the remainder: 9,999 − 35 = 9,964.
Answer: 9,964
EXAMATLAS LEVEL
Q. Find the sum of the smallest and the largest five-digit numbers that are divisible by both 7 and 11.
Divisible by 7 and 11 means divisible by 77. Smallest: 10,000 ÷ 77 leaves remainder 67, so add 77 − 67 = 10 to get 10,010. Largest: 99,999 ÷ 77 leaves remainder 53, so subtract to get 99,946. Sum = 10,010 + 99,946 = 1,09,956. Note the asymmetry: for the smallest you add (divisor − remainder), for the largest you subtract the remainder.
Answer: 1,09,956
Pattern 6: Least number to add or subtract
[PYQ: NTPC UG CBT-1 12-Aug-2025 Shift-3 | NTPC 2020-21 cycle]
EXAM LEVEL
Q. What least number must be subtracted from 10,000 to make it divisible by 23?
23 × 434 = 9,982, so 10,000 leaves remainder 18. Subtract 18.
Answer: 18
EXAMATLAS LEVEL
Q. What least number must be added to 8,47,325 so that the sum is divisible by 11?
Count places from the right because the added number changes the unit digit. Odd places from the right: 5 + 3 + 4 = 12. Even places: 2 + 7 + 8 = 17. Odd minus even = −5, so the number is 5 short of a multiple of 11 on the odd-place side. Adding 5 raises the unit digit (an odd place) by 5 and the difference becomes 0: 8,47,330 = 11 × 77,030. Counting from the left gives +5 instead (the sign flips because the number has six digits). Students who read that +5 as the remainder add 11 − 5 = 6, which is wrong; the true remainder of 8,47,325 ÷ 11 is 6, so 5 must be added.
Answer: 5
Pattern 7: Algebraic divisibility (aⁿ ± bⁿ and n³ − n)
[PYQ: NTPC CBT-1 5-Apr-2021 Shift-2 | NTPC 2020-21 cycle]
EXAM LEVEL
Q. (47⁴³ + 43⁴³) is always divisible by which of the following? (a) 4 (b) 47 (c) 90 (d) 43
aⁿ + bⁿ with n odd is divisible by a + b = 90.
Answer: (c) 90
EXAMATLAS LEVEL
Q. 5⁸ − 1 is NOT divisible by which of the following? (a) 13 (b) 24 (c) 31 (d) 39
5⁸ − 1 = (5⁴ − 1)(5⁴ + 1) = 624 × 626. 624 = 2⁴ × 3 × 13, so 13, 24 = 8 × 3 and 39 = 3 × 13 all divide it. 626 = 2 × 313 adds nothing new. For 31: 5³ = 125 = 4 × 31 + 1, so 5³ leaves 1, 5⁶ leaves 1 and 5⁸ = 5⁶ × 25 leaves 25; then 5⁸ − 1 leaves 24, not 0.
Answer: (c) 31
Same family, asked on 5 April 2021: n³ − n = (n − 1) × n × (n + 1) is a product of three consecutive integers, so it is divisible by 6 for every natural number n (and by 24 whenever n is odd).
60-Second Revision
- 11: odd-place sum minus even-place sum = 0 or multiple of 11; count from the right when a digit is added.
- Composite divisor: test CO-PRIME factors (72 = 8 × 9, 24 = 3 × 8).
- 1001 = 7 × 11 × 13: alternate sum of 3-digit groups; abcabc always divisible by 7, 11, 13.
- aⁿ − bⁿ by a − b always, by a + b when n even; aⁿ + bⁿ by a + b when n odd.
- Count of multiples of k up to N = ⌊N/k⌋; "4 or 6 but not both" = A + B − 2(both).
- Smallest n-digit multiple: add (divisor − remainder); largest: subtract remainder.