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Find the remainder when 73 × 75 × 78 × 57 × 197 × 37 is divided by 34.
Answer & Solution
Correct Answer:
Option
A
Remainder,
$$\frac{{73 \times 75 \times 78 \times 57 \times 197 \times 37}}{{34}}$$
$$ = \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$$
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
$$\eqalign{ & \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}} \cr & = \frac{{35 \times 30 \times 23 \times 27}}{{34}} \cr & = \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr} $$
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]
$$\eqalign{ & \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr & = \frac{{28 \times - 11}}{{34}} \cr & = \frac{{ - 6 \times - 11}}{{34}} \cr & = \frac{{66}}{{34}} \cr & {\text{R}} = 32 \cr} $$
Required remainder = 32
$$\frac{{73 \times 75 \times 78 \times 57 \times 197 \times 37}}{{34}}$$
$$ = \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}}$$
[We have taken individual remainder, which means if 73 is divided by 34 individually, it will give remainder 5, 75 divided 34 gives remainder 7 and so on.]
$$\eqalign{ & \frac{{5 \times 7 \times 10 \times 23 \times 27 \times 3}}{{34}} \cr & = \frac{{35 \times 30 \times 23 \times 27}}{{34}} \cr & = \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr} $$
[We have taken here negative as well as positive remainder at the same time. When 30 divided by 34 it will give either positive remainder 30 or negative remainder -4. We can use any one of negative or positive remainder at any time.]
$$\eqalign{ & \frac{{1 \times - 4 \times - 11 \times - 7}}{{34}} \cr & = \frac{{28 \times - 11}}{{34}} \cr & = \frac{{ - 6 \times - 11}}{{34}} \cr & = \frac{{66}}{{34}} \cr & {\text{R}} = 32 \cr} $$
Required remainder = 32
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LoginWe will break 67 , so that we can get multiple of 7
I.e.
63+4^99÷7 = 4^99÷7
(4^3)^33÷7= (64)^33÷7
1^33÷7= 1 remainder
Divide each number by 34.. e.g., 73/34, R(reminder)=5.. like that, u will get... (5*7*10*23*27*3)/34 = 35*23*27*30 / 34. Now 35/34, R=1 and {{23/34 (or) also = - (34/23)=R= - 11}} & -(34/27)=R= -(7).
(1*(-4)*(-11)*(-7)) / 34 = -(-28)*(-11) /34= [28*(-11)] / 34=
[(-6)*(-11)]/34 (since, -[(34/28)]=-6) so, finally 66/34= R=32, Q=1._ _
I don't understood how 30 came here?