Last digit of 4^1 = 4 ; 4^3 = 4 ; 4^5= 4....
So, 99 = 1+(n-1)*2 n = 50 i.e. 4+4+4....50 times i.e. 4*50 = 200 (last digit 0)
Similarly, last digit of 9^2=1 ; 9^4 = 1....
So, 100 = 2+(n-1)*2 n=50 i.e. 9+9+9+....50 times i.e. 9*50 = 450 (last digit 0)
So, 200+450 = 650 i.e. last digit is '0'.
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A number 2088 by what if we divide the number it will become a perfect square?
The digital sum of number is
(1+2+3+4+5+6+7+8+9+0+1+2+3+4+5+6+7+8+9)=90=(9+0) = 9
So it is divisible by 3 or 9
The given divisor 6561 = 9^4
So, 9^24/9^4 Reminder is 0
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Let a, b, c, d and e be distinct integers in ascending order such that(76-a)(76-b)(76-c)(76-d)(76-e) = 1127. What is a + b + c + d
Product of 5 terms equal to 1127. As all the five terms are integers, given product should be a product of 5 numbers. Now factorize 1127.
1127 = 72 * 23 = 7 * 7 * 23
But given that all the a, b, c, d, e are distinct. And we are getting only 3 terms with 7 repeats.
Now the logic is, integers means positive and negative, 7 and - 7 possible and 1, - 1 also possible . As a,b, c, d, e are in ascending order, the factors should be in decreasing order. So (23, 7, 1, -1, -7)
Now a = 53; b = 69; c = 75; d = 77
a + b + c + d = 274.
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For all integral values of n, the expression ((7^2n)-(3^3n)) is a multiple of:
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