Binomial Theorem And Its Simple Applications Question 92
Question: The minimum positive integral value of m such that $ {{(1073)}^{71}}-m $ may be divisible by 10, is
Options:
A) 1
B) 3
C) 7
D) 9
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ {{(1073)}^{71}}-m={{(73+1000)}^{71}}-m $
$ ={{,}^{71}}C_0{{(73)}^{71}}+{{,}^{71}}C_1{{(73)}^{70}}(1000)+{{,}^{71}}C_2{{(73)}^{69}} $
$ {{(1000)}^{2}}+…..+{{,}^{71}}C_{71}{{(1000)}^{71}}-m $
Above will be divisible by 10 if $ ^{71}C_0{{(73)}^{71}} $ is divisible by 10
Now $ ^{71}C_0{{(73)}^{71}}={{(73)}^{70}}.73={{(73^{2})}^{35}}.73 $
The last digit of $ 73^{2} $ is 9, so the last digit of $ {{( 73^{2} )}^{35}} $ is 9.
$ \therefore $ Last digit of $ {{( 73^{2} )}^{35}}.73 $ is 7
Hence, the minimum positive integral value of m is 7, so that is divisible by 10.
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