Barring any calculation errors, here's how to do it manually:

Find x, where:

x = 397^611 mod 1763

Step 1.

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Note 1763 = 42^2 - 1 = 41*43.

Step 2.

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You now have 2 problems:

Find a, where:

a = 397^611 mod 41, and

Find b, where:

b = 397^611 mod 43.

Step 3:

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Note: 397 == -13 (41), so

a = (-13)^611 (41).

Use Fermat's little theorem to say for any y, y^40 == 1 (41) and

note 611 == 11 (40), so

a = (-13)^11 (41)

a = 169^5 * 28 (41)

a = 5^5 * 28 (41)

a = 125 * 25 * 28 (41)

a = 2 * 25 * 28 (41)

a = 15 * 25 (41)

a = 375 (41)

a = 6 (41)

Step 4:

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Note 397 == 10 (43), and 611 == 23 (42), as before:

b = 10 ^ 23 (43)

b = 1000 ^ 7 * 100 (43)

b = 1000 ^ 7 * 100 (43)

b = 11 ^ 7 * 14 (43)

b = 121 ^ 3 * 154 (43)

b = (-2) ^ 3 * 25 (43)

b = -200 (43)

b = 15(43)

Step 5:

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Since 2*21==1(41), 43*21==1(41).

Since (-2)*(-21)==1(43), 41*21==1(43)

The number

n = 6*43*21 + 15*41*21 satisfies both congruences in step 2.

The Chinese remainder theorem says it also equals x.

x = 21*(6*43+15*41) mod (41*43)

x = 703 + 1763*k, with k any integer.

QED