Last three digits of 17256 is

Question

Last three digits of 17256 is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Last three digits of 17256 =(289)128=(- 1 + 290)128 =128C0(- 1)128-128C1⋅ 290+128C2⋅ (290)2+....... =1-128× 290+64× 127× 292× 100+....... =1-......120+.....800+....... =681

Q. Last three digits of $1 7^{256}$ is

Last three digits of $1 7^{256}$
$= (289)^{128} = (- 1 + 290)^{128}$
$=^{128} C_{0} (- 1)^{128} -^{128} C_{1} \cdot 290 +^{128} C_{2} \cdot (290)^{2} + .......$
$= 1 - 128 \times 290 + 64 \times 127 \times 2 9^{2} \times 100 + .......$
$= 1 - ......120 + .....800 + .......$
$= 681$

Q. Last three digits of 17256 is

Last three digits of 17256 =(289)128=(−1+290)128 =128C0​(−1)128−128C1​⋅290+128C2​⋅(290)2+....... =1−128×290+64×127×292×100+....... =1−......120+.....800+....... =681

Q. Last three digits of $1 7^{256}$ is

Last three digits of $1 7^{256}$
$= (289)^{128} = (- 1 + 290)^{128}$
$=^{128} C_{0} (- 1)^{128} -^{128} C_{1} \cdot 290 +^{128} C_{2} \cdot (290)^{2} + .......$
$= 1 - 128 \times 290 + 64 \times 127 \times 2 9^{2} \times 100 + .......$
$= 1 - ......120 + .....800 + .......$
$= 681$