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Re: If x is a positive integer, what is the units digit of
[#permalink]
18 Sep 2017, 02:20
1
Carcass wrote:
If x is a positive integer, what is the units digit of \((24)^{5+2x}\) + \((36)^6\) + \((17)^3\)
(A) 2 (B) 3 (C) 4 (D) 6 (E) 8
a n
Any hint for this one? I computed 36^6 which terminates with a 6, 17^3 terminates with a 3 and 24^(5+2x) can be rewritten as 24^5*24^2x where 24^5 terminates with a 4 and 24^2x terminates with a 6 for whatever value of x. Thus, the 24 term terminates with a 4. Summing three numbers terminating with 6, 3 and 4, it results a number terminating with 3, thus answer B seems right to me. However, the answer seems to be A. why?
If x is a positive integer, what is the units digit of
[#permalink]
23 May 2022, 10:41
1
We need to find the units digit of \((24)^{5+2x}\) + \((36)^6\) + \((17)^3\)
Now, Units digit of \((24)^{5+2x}\) will be same as units digit of \((4)^{5+2x}\) (As units digit of 24 is same as 4 which is 4) Units digit of \((36)^6\) will be same as units digit of \((6)^6\) (As units digit of 36 is same as 6 which is 6) Units digit of \((17)^3\) will be same as units digit of \((7)^3\) (As units digit of 17 is same as 7 which is 7)
Let's find the units digit of \((4)^{5+2x}\)
To find the units digit of power of 4 we need to check the cyclicity in the units digit of powers of 4
\(4^1\) units’ digit is 4 [ 4 ] \(4^2\) units’ digit is 6 [ 16 ] \(4^3\) units’ digit is 4 [ 64 ] \(4^4\) units’ digit is 6 [ 256 ]
=> Units digit of any positive odd integer power of 4 is 4 => Units digit of any positive even integer power of 4 is 6 => Now 2x + 5 will be Evan + Odd = Odd (As 2x is Even and 5 is odd) => Units digit of \((4)^{5+2x}\) = 4
Let's find the units digit of \((6)^6\)
To find the units digit of power of 6 we need to check the cyclicity in the units digit of powers of 6
\(6^1\) units’ digit is 6 [ 6 ] \(6^2\) units’ digit is 3 [ 36 ] \(6^3\) units’ digit is 3 [ 216 ]
=> Units digit of any positive integer power of 6 is 6 => The units digit of \(6^6\) = 6
Let's find the units digit of \((7)^3\)
\(7^1\) units’ digit is 7 [ 7 ] \(7^2\) units’ digit is 9 [ 7*7 = 49 ] \(7^3\) units’ digit is 3 [ 9*7 = 63 ] (P.S>: we don't need to proceed further) \(7^4\) units’ digit is 1 [ 3*7 = 21 ] \(7^5\) units’ digit is 3 [ 1*7 = 7 ]
=> Units digit of power of 7 repeats after every \(4^th\) number
=> Units digit of \((24)^{5+2x}\) + \((36)^6\) + \((17)^3\) = Units digit of 4 + 6 + 3 = Units digit of 13 = 3
So, Answer will be B. Hope it helps!
Watch the following video (from 2:48 mins) to learn how to find cyclicity of 3 and other numbers