Number System — Practice Questions
CSAT-pattern number system practice — divisibility by 11, HCF–LCM pairs, power remainders, traffic-light LCM, divisor-of-divisor remainders and number properties, fully worked.
Questions
These are original CSAT-pattern practice questions written for this site — not past UPSC papers. No calculator; target time about 9 minutes.
Consider the following numbers:
- 2,794
- 30,251
- 4,785
- 7,392
How many of the above numbers are divisible by 11?
Solution
Alternating digit sum, starting from the units digit:
| Number | Alternating sum | Divisible by 11? |
|---|---|---|
| 2,794 | $4 - 9 + 7 - 2 = 0$ | Yes |
| 30,251 | $1 - 5 + 2 - 0 + 3 = 1$ | No |
| 4,785 | $5 - 8 + 7 - 4 = 0$ | Yes |
| 7,392 | $2 - 9 + 3 - 7 = -11$ | Yes |
Three numbers.
Answer: C
The product of two numbers is 2,160 and their HCF is 12. How many such pairs of numbers are possible?
Solution
Let the numbers be $12a$ and $12b$ with $a, b$ co-prime. Then $144ab = 2{,}160$, so $ab = 15$.
Co-prime pairs with product 15: $(1, 15)$ and $(3, 5)$ → numbers $(12, 180)$ and $(36, 60)$.
Check: $\text{HCF}(36, 60) = 12$ and $36 \times 60 = 2{,}160$ ✓.
Answer: B
What is the remainder when $2^{31}$ is divided by 5?
Solution
Remainders of $2^1, 2^2, 2^3, 2^4$ on division by 5: $2, 4, 3, 1$ — then the cycle repeats.
$31 = 4 \times 7 + 3$ → third term → 3.
Answer: B
Three traffic lights change every 48 seconds, 72 seconds and 108 seconds respectively. If they change together at 8:20:00 a.m., when will they next change together?
Solution
$48 = 2^4 \times 3$, $72 = 2^3 \times 3^2$, $108 = 2^2 \times 3^3$.
$\text{LCM} = 2^4 \times 3^3 = 16 \times 27 = 432$ seconds $= 7$ min $12$ s.
$8{:}20{:}00 + 7{:}12 = $ 8:27:12 a.m.
Answer: A
A number leaves a remainder of 39 when divided by 357. What is the remainder when the same number is divided by 17?
Solution
$357 = 17 \times 21$, so 17 divides 357. The number is $357k + 39$; the part $357k$ is a multiple of 17.
Remainder $= 39 \bmod 17 = 39 - 34 = 5$.
Answer: D
Consider the following statements:
- The product of any three consecutive natural numbers is divisible by 6.
- The square of any odd number leaves a remainder of 1 when divided by 8.
- Every prime number greater than 3 is of the form $6k + 1$ or $6k - 1$.
Which of the statements given above are correct?
Solution
- 1 — correct. Among three consecutive numbers, at least one is even and exactly one is a multiple of 3.
- 2 — correct. $(2m + 1)^2 = 4m(m+1) + 1$; $m(m+1)$ is even, so $4m(m+1)$ is a multiple of 8.
- 3 — correct. Any integer is $6k, 6k+1, \ldots, 6k+5$. The forms $6k, 6k+2, 6k+3, 6k+4$ are divisible by 2 or 3, so a prime above 3 must be $6k + 1$ or $6k + 5$ ($= 6(k+1) - 1$).
Answer: D