Divisibility is about asking one question: does this number divide evenly, leaving zero remainder? That is it. No long division needed.
Here is the analogy that makes it stick. Imagine you are splitting a pile of mangoes equally among friends. If every mango goes somewhere with nothing left over, the pile was "divisible" by the number of friends. If even one mango is left, it is not divisible.
In SSC MTS, you will rarely need to actually divide. Instead, you use digit-pattern rules — shortcuts that let you test divisibility in under five seconds by looking at the digits of the number, not computing the full quotient.
Why do these rules work? Because our number system is base-10. Each digit position carries a power of 10. The divisibility properties of those powers of 10 are what the rules exploit. You do not need to understand the derivation to use the rules — but knowing the "why" helps you remember them without rote learning.
Two more terms you need clear:
12 ÷ 4 = 3 exactly, then 4 is a factor (divisor) of 12.For SSC MTS specifically, divisibility questions appear in two flavors: (1) identify which number in the options is or is not divisible by something, and (2) find what smallest number to add or subtract from a given number to make it divisible. The second type uses the remainder concept directly. Know both patterns cold.
Here are the rules for all divisors SSC MTS has ever tested, with the logic attached so you remember them rather than memorize them blind.
Divisibility by 2
Last digit is 0, 2, 4, 6, or 8.
Logic: 10 is divisible by 2, so only the last digit matters.
Divisibility by 3
Sum of all digits is divisible by 3.
Example: 4128 → 4+1+2+8 = 15 → 15 ÷ 3 = 5 ✓
Divisibility by 4
Last two digits form a number divisible by 4.
Example: 4732 → last two digits 32 → 32 ÷ 4 = 8 ✓
Logic: 100 is divisible by 4, so only the last two digits matter.
Divisibility by 5 Last digit is 0 or 5.
Divisibility by 6 Divisible by both 2 AND 3. (Both rules must pass.)
Divisibility by 7
No clean digit rule. Divide directly or use the "subtract twice the last digit" method:
Take the number, subtract twice its last digit, check if the result is divisible by 7. Repeat if needed.
Example: 8956 → last digit 6, twice = 12 → 895 − 12 = 883 → last digit 3, twice = 6 → 88 − 6 = 82 → 82 ÷ 7 is not exact → remainder exists.
For SSC MTS, dividing directly is often faster for 7.
Divisibility by 8
Last three digits form a number divisible by 8.
Example: 4128 → last three digits 128 → 128 ÷ 8 = 16 ✓
Logic: 1000 is divisible by 8, so only the last three digits matter.
Divisibility by 9
Sum of all digits is divisible by 9.
Example: 6804 → 6+8+0+4 = 18 → 18 ÷ 9 = 2 ✓
This is the most-tested rule in SSC MTS. Learn it first.
Divisibility by 10 Last digit is 0.
Divisibility by 11
Alternating sum rule: (sum of digits at odd positions) − (sum of digits at even positions) must be 0 or a multiple of 11.
Position counting starts from the right.
Example: 4543
Odd positions from right: 3, 5, 4 → sum = 12
Even positions from right: 4 → sum = 4
Wait — let's count carefully. 4543: positions right to left are 3(pos 1), 4(pos 2), 5(pos 3), 4(pos 4).
Odd positions: 3 + 5 = 8. Even positions: 4 + 4 = 8. Difference = 0. Divisible by 11. ✓
Divisibility by 12 Divisible by both 3 AND 4.
Divisibility by 13
No standard digit rule. Direct division, or use the fact that 7 × 11 × 13 = 1001 — any number of the form abc abc (six digits, first three = last three) is divisible by 1001, hence by 7, 11, and 13.
This is where SSC MTS takes divisibility one step further. Look —
Pattern: "What least number should be subtracted from N to make it divisible by D?"
Method:
D − R.Example: Subtract from 8956 to make divisible by 7.
8956 ÷ 7 = 1279 remainder 3.
Subtract 3: 8956 − 3 = 8953 = 7 × 1279 ✓
Example: Add to 4537 to make divisible by 11.
For 11, use the alternating sum: (4−5+3−7) = −5.
Distance from −5 to 0 (nearest multiple of 11) is 5. But adding 5 to the number changes the last digit: 4537 + 5 = 4542. Check: (4−5+4−2) = 1. Not 0. This is why for 11, you must recheck after adding.
Correct approach: just find 4537 mod 11 by actual division. 4537 ÷ 11 = 412 remainder 5. So you need 11 − 5 = 6. Add 6: 4543. 4543 ÷ 11 = 413. ✓
When a question asks "divisible by both X and Y," check both rules independently. The number must pass both tests.
If X and Y share no common factor (i.e., they are co-prime), you can also treat it as: the number must be divisible by X × Y.
For any divisor of 9 questions, remember: the digit sum of N, when divided by 9, gives you the same remainder as N itself divided by 9. So if digit sum = 17, remainder = 17 mod 9 = 8. You never need to divide the original large number. Standard method on 6812: long division by 9 takes ~30 seconds. Digit-sum method: 6+8+1+2=17, 17−9=8, done in 8 seconds.
Divisibility by 4 → check last 2 digits only. Divisibility by 8 → check last 3 digits only. The pattern: the number of digits you check equals the power of 2 in the divisor (4=2², check 2 digits; 8=2³, check 3 digits). When you see a 4-digit or 5-digit number, mentally chop off everything except the last 2 or 3 digits and divide those small numbers. Checking divisibility of 4951 by 4: ignore 49, just check 51 ÷ 4 = 12.75 → not divisible. Time: 4 seconds vs. full division in 20+ seconds.
Questions asking "divisible by both X and Y" have four options. Apply the easier/faster rule first to eliminate. Example: "divisible by both 3 and 8." Rule for 8 (last 3 digits) is faster than digit sum. Check last 3 digits of each option first — eliminate any that fail. Then apply the digit-sum rule only to survivors. In a 4-option question, you often eliminate 3 options with a single rule, making the second rule check unnecessary. This cuts solving time from ~40 seconds to ~15 seconds.
Any number divisible by 1001 is automatically divisible by 7, 11, and 13 simultaneously. Why? 1001 = 7 × 11 × 13 (all prime, so LCM = product). So 1001, 2002, 3003 ... are all divisible by all three. Also, any 6-digit number of form ABCABC equals ABC × 1001, so it is divisible by 7, 11, and 13. Recognizing 1001 instantly in options eliminates the need to test 7, 11, 13 separately — saving 3 separate divisions.
Find R = N mod D (divide N by D, take remainder). To make N divisible by D: subtract R (if R ≠ 0); add (D − R). Never subtract 0 — if R = 0, the number is already divisible. This two-line framework applies to any divisor (3, 7, 9, 11...) without separate memorization. Standard approach for each divisor type as a separate case: ~45 seconds of confusion. This unified rule: ~15 seconds once you have the remainder.
In the exam hall, use this decision tree:
Step 1 — Read the question type.
Step 2 — Pick the right rule.
Step 3 — Cross-check with one option only. Once you identify the answer, verify it satisfies the condition. Do not verify all four options — time wasted.
Why this question: The classic digit-sum remainder question for divisibility by 9. This exact pattern has appeared multiple times across SSC papers.
Solving path: Digit sum of 6812 = 6+8+1+2 = 17. 17 mod 9 = 8 (since 17 − 9 = 8). So the remainder when 6812 is divided by 9 is 8. Subtracting 8 gives remainder 0. Answer: 8.
Why this question: Tests pure rule recall for divisibility by 4. The trap is that students sometimes check only the last digit instead of the last two digits.
Solving path: Extract last two digits from each option — 32, 12, 44, 51. Check each: 32÷4=8 ✓, 12÷4=3 ✓, 44÷4=11 ✓, 51÷4=12.75 ✗. The one that fails is 4951. Answer: 4951.
Why this question: Double-condition question — divisible by both 3 and 4. Tests whether you can apply two rules in sequence and use elimination.
Solving path: Start with rule for 4 (faster). Last two digits: 10320→20 (20÷4=5 ✓), 8235→35 (35÷4=8.75 ✗), 8491→91 (91÷4=22.75 ✗), 7582→82 (82÷4=20.5 ✗). Only 10320 survives. Confirm rule for 3: 1+0+3+2+0=6, divisible by 3 ✓. Answer: 10320.
Why this question: Combines divisibility by 3 (digit sum) and divisibility by 8 (last three digits). Good example of elimination strategy.
Solving path: Rule for 8 first — last three digits: 4136→136 (136÷8=17 ✓), 4112→112 (112÷8=14 ✓), 4128→128 (128÷8=16 ✓), 4192→192 (192÷8=24 ✓). All pass rule for 8. Now rule for 3 — digit sums: 4+1+3+6=14 (not divisible by 3 ✗), 4+1+1+2=8 ✗, 4+1+2+8=15 (divisible by 3 ✓), 4+1+9+2=16 ✗. Only 4128 passes both. Answer: 4128.
Why this question: Tests the combined "neither...nor" condition — you must verify both rules fail. Requires checking all options methodically.
Solving path: Check divisibility by 4 (last two digits) and 9 (digit sum) for each option. 3312: last two = 12 (÷4=3 ✓, so divisible by 4 — eliminated). 2363: last two = 63 (63÷4=15.75 ✗); digit sum = 2+3+6+3=14 (14÷9 not exact ✗). Both rules fail → this is the answer. No need to check 3492 or 3528. Answer: 2363.
Why this question: The 7 × 11 × 13 = 1001 pattern question. Rewards those who know the product shortcut — otherwise you are doing three separate divisions.
Solving path: LCM(7, 11, 13) = 7 × 11 × 13 = 1001 (all three are prime, no common factors). So the smallest number divisible by all three is 1001. Check options: 1001 is directly the LCM. Verify: 1001÷7=143 ✓, 1001÷11=91 ✓, 1001÷13=77 ✓. Answer: 1001.
Why this question: "Least number to add" with divisibility by 11 — combines the alternating sum rule with the add-formula. A common error is applying the alternating sum incorrectly.
Solving path: Find 4537 mod 11. Direct division: 4537 ÷ 11 = 412 remainder 5. To find the least number to add: 11 − 5 = 6. Check: 4537 + 6 = 4543. 4543 ÷ 11 = 413 ✓. Answer: 6.
Why this question: "Least number to subtract" for divisibility by 7. Straightforward remainder application, but 7 division trips students who try to apply digit-sum rules (which do not exist for 7).
Solving path: 8956 ÷ 7: 7 × 1279 = 8953, so 8956 − 8953 = 3. Remainder = 3. Subtract 3 from 8956: 8953 ÷ 7 = 1279 ✓. Answer: 3.
Checking only the last digit for divisibility by 4. The rule requires the last two digits. A number ending in 2 could still be divisible by 4 (e.g., 412: 12÷4=3 ✓), or a number ending in 8 might not be (e.g., 418: 18÷4=4.5 ✗). Always take two digits.
Confusing digit sum rules for 3 and 9. Both use digit sum, but divisibility by 9 requires the digit sum itself to be divisible by 9 (not just by 3). Digit sum of 15 → divisible by 3 but not 9. Digit sum of 18 → divisible by both.
Applying rule for 8 to only two digits. Divisibility by 8 needs the last three digits, not two. Checking 4128: do not look at just 28 (28÷8=3.5 ✗); look at 128 (128÷8=16 ✓). The correct answer flips.
For alternating sum (divisibility by 11), miscounting positions. Position 1 is the rightmost digit. Count from the right. A common error is counting from the left and getting the wrong sign for the difference.
Using remainder to subtract when the question asks to add, and vice versa. Read the question again. "Subtract" → answer is R. "Add" → answer is D − R. These are different and the options will include both as traps.
Assuming combined divisibility means you multiply the divisors blindly. Divisibility by both 4 and 6 does NOT mean divisibility by 24. You must take the LCM: LCM(4,6)=12, not 24. Multiplying works only when the two divisors are co-prime (share no common factors).