# Remainder of a to the power n, divided by b. | Algebra - Number Theory

**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 7^{64}/8.

Answer: 1

**Explanation** :

Let us find the pattern that remainders follow when successive powers of 7 are divided by 8.

Remainder of 7^{1}/8 = 7.

Remainder of 7^{2}/8 = 1.

Remainder of 7^{3}/8 = 7.

Remainder of 7^{4}/8 = 1.

∴ We find that the remainders are repeated after every two powers.

∴ Remainder of 7^{64} when divided by 8 is the same as 7^{2} when divided by 8, since 64 is a multiple of 2.

⇒ The remainder is 1.

**Alternately,**

$R\left[\frac{{7}^{64}}{8}\right]$ = ${\left[R\left(\frac{7}{8}\right)\right]}^{64}$ = [-1]^{64} = 1

Hence, 1.

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**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 2^{100}/7.

Answer: 2

**Explanation** :

Let us find the pattern that remainders follow when successive powers of 2 are divided by 7.

Remainder of 2^{1}/7 = 2.

Remainder of 2^{2}/7 = 4.

Remainder of 2^{3}/7 = 1.

Remainder of 2^{4}/7 = 2.

Remainder of 2^{5}/7 = 4.

Remainder of 2^{6}/7 = 1.

∴ We find that the remainders are repeated after every three powers.

∴ $R\left[\frac{{2}^{100}}{7}\right]$ = $R\left[\frac{{2}^{99}\times {2}^{1}}{7}\right]$ = $R\left[\frac{{2}^{99}}{7}\right]\times R\left[\frac{{2}^{1}}{7}\right]$ = 1 × 2 = 2

⇒ The remainder is 2.

Hence, 2.

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**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 3^{100}/7.

Answer: 4

**Explanation** :

Let us find the pattern that remainders follow when successive powers of 3 are divided by 7.

Remainder of 3^{1}/7 = 3.

Remainder of 3^{2}/7 = 2.

Remainder of 3^{3}/7 = 6.

Remainder of 3^{4}/7 = 4.

Remainder of 3^{5}/7 = 5.

Remainder of 3^{6}/7 = 1.

Remainder of 3^{7}/7 = 3.

Remainder of 3^{8}/7 = 2.

∴ We find that the remainders are repeated after every six powers.

∴ $R\left[\frac{{3}^{100}}{7}\right]$ = $R\left[\frac{{3}^{96}\times {3}^{4}}{7}\right]$ = $R\left[\frac{{3}^{96}}{7}\right]\times R\left[\frac{{3}^{4}}{7}\right]$ = 1 × 4 = 4.

⇒ The remainder is 4.

**Alternately,**

Let us find the pattern that remainders follow when successive powers of 3 are divided by 7.

Remainder of 3^{1}/7 = 3.

Remainder of 3^{2}/7 = 2.

Remainder of 3^{3}/7 = 6 = -1. (Concept of negative remainder)

∴ R(3^{3}/7) = -1

∴ $R\left[\frac{{3}^{100}}{7}\right]$= $R\left[\frac{{3}^{99}\times {3}^{1}}{7}\right]$ = $R\left[\frac{{3}^{99}}{7}\right]\times R\left[\frac{{3}^{1}}{7}\right]$ = ${\left(R\left[\frac{{3}^{3}}{7}\right]\right)}^{33}$ × 3 = (-1)^{33} × 3 = -3 ≡ 7 - 3 = 4.

Hence, 4.

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**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 7^{83} divided by 10.

Answer: 3

**Explanation** :

Let us find the pattern that remainders follow when successive powers of 3 are divided by 7.

Remainder when 7^{1}/10 = 7.

Remainder when 7^{2}/10 = 9.

Remainder when 7^{3}/10 = 3.

Remainder when 7^{4}/10 = 1.

Remainder when 7^{5}/10 = 7.

Remainder when 7^{6}/10 = 9.

∴ We find that the remainders are repeated after every four powers.

$R\left[\frac{{7}^{83}}{10}\right]$ = $R\left[\frac{{7}^{80}\times {7}^{3}}{10}\right]$ = $R\left[\frac{{7}^{80}}{10}\right]$ × $R\left[\frac{{7}^{3}}{10}\right]$ = 1 × 3 = 3.

**Alternately,**

Remainder of any number when divided by 10 is same as the last digit of that number.

∴ We have to find the last digit of 7^{83}.

Last of 7^{83} = last digit of 7^{80} × 7^{3} (Cyclicity of last digit of powers of 7 is 4.)

= last digit of 7^{3}

= 3

Hence, 3.

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**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 6^{33} divided by 64.

Answer: 0

**Explanation** :

There is no need to find the pattern of remainders here.

Divisor, 64 = 2^{6}.

Power of 2 in 6^{33} is definitely greater than 6.

∴ 6^{33} is perfectly divisible by 64 i.e., remainder is 0.

Hence, 0.

Workspace:

**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 2^{120}/17.

Answer: 1

**Explanation** :

Let us find the pattern that remainders follow when successive powers of 2 are divided by 17.

Remainder of 2^{1}/17 = 2.

Remainder of 2^{2}/17 = 4.

Remainder of 2^{3}/17 = 8.

Remainder of 2^{4}/17 = 16 = -1.

∴ Remainder will repeat after every 8 terms.

or, R[2^{4}/17] = -1

⇒ R[2^{120}/7] = R[2^{4×30}/7] = R[2^{4}/7]^{30} = (-1)^{30} = 1.

Hence, 1.

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**Remainder of a to the power n, divided by b. | Algebra - Number Theory**

Find the remainder of the division 19^{122}/17.

Answer: 4

**Explanation** :

R[19^{122}/17] = R[19/17]^{122} = R[2^{122}/17]

Let us find the pattern that remainders follow when successive powers of 2 are divided by 17.

Remainder of 2^{1}/17 = 2.

Remainder of 2^{2}/17 = 4.

Remainder of 2^{3}/17 = 8.

Remainder of 2^{4}/17 = 16 = -1.

∴ Remainder will repeat after every 8 terms.

or, R[2^{4}/17] = -1

⇒ R[2^{122}/7] = R[2^{4×30+2}/7] = R[2^{120}/7] × R[2^{2}/7] = R[2^{4}/7]^{30} × R[2^{2}/7] = (-1)^{30} × 4 = 4.

Hence, 4.

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