Question

${2}^{60} when \: divided \: by \: 7 \: leaves \: the \: remainder. \: please \: solve \: this \: question \: and \: explain \: this$
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Answer 1

2 divided by 7 leaves remainder 2. This can be written as the following:

2 ≡ 2 (mod 7)

Cubing both sides, we get,

2³ ≡ 2³ (mod 7)

8 ≡ 8 (mod 7)

8 leaves remainder 1 on division by 7. Thus,

8 ≡ 1 (mod 7)

∴ 2³ ≡ 1 (mod 7)

Now we can find the answer because 1 is got as remainder here.

Take the 20th power of each, we get,

(2³)²⁰ ≡ 1²⁰ (mod 7)

2⁶⁰ ≡ 1 (mod 7)

This means that 2⁶⁰ divided by 7 leaves remainder 1.

Thus, the answer is 1.

Answer 2

$\huge{\underline{\underline{Answer\::-}}}$

♦ We have to find out the reminder when $2^{60}$ is divided by 7 .

♦ To find we must find out the least power of 2 to which we can divided 7 .

♦ Condition required

• It should be equal or greater than 7 .

♦ Now According to the condition the number which satisfies it is $2^3$ or 8

♦ Now we will find out the reminder when 8 is divided by 7

= 8 ÷ 7

= 7(1) + 1

So reminder is 1

♦ Now we will check it with other power of 8

$8^2$ ÷ 7

= 64 ÷ 7

= 7(9) + 1 {same reminder}

$8^3$ ÷ 7

= 512 ÷ 7

= 7(73) + 1 { same reminder }

♦ So by above's results we can conclude that

$8^x$ ÷ 7 = 7n + 1

♦ Now coming back to the question .

• We can write $2^{60}$

as ${( 2^3 )}^{20}$

Or ${8}^{20}$

Then from above's conclusions that

$8^x$ ÷ 7 = 7n + 1

We can say that

$8^{20}$ ÷ 7 = 7n + 1

So, Reminder = 1