0

Find the sum of last three digits of the number $89^{23}$ I arrived to calculate this expression by taking mod 1000.

$89^{23}$ $\mod{1000}$ but what to do next .i stopped . help to sort out this.

Riyas Randhava
  • 55
  • See http://math.stackexchange.com/questions/519838/last-three-digits-of-23320, http://math.stackexchange.com/questions/108868/find-the-last-3-digits-of-the-number-200320022001, https://www.quora.com/What-are-the-last-three-digits-of-107-107 – lab bhattacharjee Nov 26 '16 at 13:08
  • You should show your attempts to get a reasonable answer here. – StubbornAtom Nov 26 '16 at 13:14
  • I m able to get only last two digit like this $89^{23}$ = $9^{23}$ = ${{9^{2}}^{11}} 9$ =$81^{11}9$=81*9=29 and unable to get last 3. – Riyas Randhava Nov 26 '16 at 13:24

1 Answers1

2

$$89^{23}=(90-1)^{23}=-1+\binom{23}190-\binom{23}290^2\pmod{1000}$$

Now $\binom{23}29^2=23\cdot11\cdot81\equiv3\pmod{10}\implies\binom{23}290^2\equiv3\cdot100\pmod{10\cdot100}$

and $\binom{23}190=23\cdot90\equiv2070\equiv70\pmod{1000}$

So,$$89^{23}\equiv-1+70-300\pmod{1000}\equiv-1+70+700\equiv?$$

lab bhattacharjee
  • 274,582