# Find the least positive integer n that satisfies the congruence 7^128 congruent to n (mod13)

Find the least positive integer n that satisfies the congruence
${7}^{128}$ congruent to n (mod13)
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Step 1
Given,
${7}^{128}\equiv n\left(\text{mod}13\right)$
Find the value of n that satisfies the above congruence.
Step 2
${7}^{128}\equiv n\left(\text{mod}13\right)$
Since 7 is not the multiple of 13.
Therefore,
${7}^{12}\equiv 1\left(\text{mod}13\right)$
Then,
${7}^{128}\equiv n\left(\text{mod}13\right)$
$⇒{\left({7}^{12}\right)}^{10}×{7}^{8}\equiv n\left(\text{mod}13\right)$
$⇒{\left(1\right)}^{10}×{7}^{8}\equiv n\left(\text{mod}13\right)$
$⇒1×{7}^{8}\equiv n\left(\text{mod}13\right)$
$⇒{7}^{8}\equiv n\left(\text{mod}13\right)$
Step 3
Also,
${7}^{2}\equiv 10\left(\text{mod}13\right)$
So,
${7}^{8}\equiv n\left(\text{mod}13\right)$
$⇒{\left({7}^{2}\right)}^{4}\equiv n\left(\text{mod}13\right)$
$⇒{\left(10\right)}^{4}\equiv n\left(\text{mod}13\right)$
$⇒{\left({10}^{2}\right)}^{2}\equiv n\left(\text{mod}13\right)\left[\therefore {\left(10\right)}^{2}\equiv 9\left(\text{mod}13\right)\right]$
$⇒{\left(9\right)}^{2}\equiv n\left(\text{mod}13\right)$
$⇒3\equiv n\left(\text{mod}13\right)$
Step 4
Hence,
n=3