Arithmetic

Numericals

Easy

applied

Which of the following holds true?

(A) $26\equiv1(mod\ 5)$

(B) $17\equiv3(mod\ 8)$

(C) $-30\equiv-15(mod\ 3)$

(D) $39\equiv12(mod\ 7)$

Choose the correct answer from the options given below:

(A) and (B) only

(A) and (C) only

(B) and (C) only

(B) and (D) only

✅ Correct Option: 2

(A): $26-1=25$ is divisible by $5$ , true. (C): $-30-(-15)=-15$ is divisible by $3$ , true.

(B) is false since $17-3=14$ is not a multiple of $8$ , and (D) is false since $39-12=27$ is not a multiple of $7$ .

2025: 2 June Shift 1

The unit's digit of $2^{2025}$ is

2025: 15 May Shift 1

In a game of billiards, A can give B, 15 points in 60 and A can give C, 20 points in 60. How many points can B give C in a game of 90?

2025: 3 June Shift 2

The least non-negative remainder, when $5^{61}$ is divided by 7, is

2026: 25 May Shift 1

The least non-negative remainder when $3^{200}$ is divided by 8 is :

2026: 19 May Shift 1

The possible value (s) of x such that $28 \equiv x(mod\ 6)$ are :-

(A) -2

(B) 4

(D) 10

Choose the correct answer from the options given below:

2026: 30 May Shift 1

The set of values of $x$ satisfying $x\equiv12(\mod5)$ is: (Where $I$ is set of integer)

Solution