Find the remainder when 73 *75 *78 *57 *197 is divided by 34.

Let N= (73•75•78•57•197)
Asked Rem. —-> N/34
So r/34 will give us Remainder (r)

Now (73•75•78•57•197)/34

N/34 —(r)—> (5•7•10•23•27)/34

N/34 —(r)—> (350•23•27)/34

N/34 —(r)—> (10•23•27)/34

N/34 —(r)—> (270• 23)/34

N/34 —(r)—> (32•23)/34
N/34 ——> 736/34
N = 34(21) +22

Hence remainder is 22 (A)

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Property : Product of remainders = Remainder of Product

So let's find the remainder when 73, 75, 78, 59 & 197 are individually divided by 37

This is 5, 7, 10, 23 & 27

So remainder of the product 73 *75 *78 *57 *197 when divided by 37 = Product of 5*7*10*23*27

But note that 5*7*10*23*27 is definitely more than 34 whereas the final remainder should be less than 34 [Any number divided by any other number 'd'can leave remainders from 0 to (d-1) ]

So we can repeat the above steps to arrive at the final remainder which will be less than 34

5*7*10*23*27 = 35*230*27

35 leaves a remainder of 1, 230 leaves a remainder of 26, 27 leaves a remainder of 27

Therefore final remainder = remainder of (1*26*27)/34 or remainder of 702/34 which is 22

Answer is (A)

Please see https://www.youtube.com/watch?v=73lEwHM ... e=youtu.be for more
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N = 73 *75 *78 *57 *197 = (68 + 5)*(68 + 7)*(68 + 10)*(68 - 11)*(204 - 7)

All terms will be a multiple of 34 except the last term obtained by multiplying 5*7*10*(-11)*(-7)

5*7*10*(-11)*(-7) = 35 * 110*7 = (34 + 1)*(102 + 8)*7

All terms will be a multiple of 34 except the last term obtained by multiplying 1*8*7 = 56

When 56 is divided by 34, remainder is 22.

Answer (A)

73*75= (5*7) mod 34= 1 (mod 34)
78=10 mod 34
57*197= (-11*-7) mod 34=9 mod 34

73 *75 *78 *57 *197= 90 mod 34= 22 mod 34

Remainder we will get is 22

(73/34)Rof = 5

*

(75/34)Rof = 7

*

(78/34)Rof = 10

*

(57/34)Rof = 23

*

(197/34)Rof = 27
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(5*7*10*23*27) / (34) Rof = ?

(35/34)Rof = 1
*
(230/34)Rof = -8
*
(27/34)Rof = -7

1 * -8 * -7 = 56 ------> Removing the Excess Remainder of 56

56 - 34 = Remainder of 22

-A-

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