What is the smallest non-negative integer k that satisfies the modular congruence 7^{99} +… — JAMB UTME

What is the smallest non-negative integer $k$ that satisfies the modular congruence $7^{99} + k \equiv -15 \pmod{11}$ ?

The correct answer is 10.

Using Fermat's Little Theorem,

7^{10} \equiv 1 \pmod{11}

, which simplifies

7^{99} \pmod{11}

7^9 \equiv 8 \pmod{11}

. Reducing the right-hand side gives

-15 \equiv 7 \pmod{11}

. The modular equation

8 + k \equiv 7 \pmod{11}

yields

k \equiv -1 \pmod{11}

. Adding the modulus

11

gives the canonical positive remainder

10

Simplify

7^{99} \pmod{11}

using Fermat's Little Theorem.

7^{10} \equiv 1 \pmod{11}

, so

7^{99} = (7^{10})^9 \times 7^9 \equiv 1^9 \times 7^9 \equiv 7^9 \pmod{11}

Since

11

is prime and

\gcd(7, 11) = 1

, Fermat's Little Theorem allows exponent reduction modulo

10

Compute

7^9 \pmod{11}

7^1 \equiv 7

7^2 = 49 \equiv 5

7^4 \equiv 5^2 = 25 \equiv 3

7^5 \equiv 7 \times 3 = 21 \equiv 10

7^9 = 7^5 \times 7^4 \equiv 10 \times 3 = 30 \equiv 8 \pmod{11}

Repeated squaring and modular multiplication efficiently reduces

7^9

modulo

11

Reduce the right-hand side

-15 \pmod{11}

-15 = -2(11) + 7 \equiv 7 \pmod{11}

Converting negative numbers into the standard non-negative remainder range

[0, 10]

Substitute remainders into the congruence and solve for

k

8 + k \equiv 7 \implies k \equiv 7 - 8 = -1 \pmod{11}

Linear algebraic rearrangement in modular arithmetic.

Convert the negative remainder

-1

to canonical non-negative form.

k = -1 + 11 = 10

The standard remainder must satisfy

0 \leq k < 11

Modular Exponentiation & Negative Remainder Reduction

Question