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20 Jan 2022, 05:15
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:33) correct
30%
(01:53)
wrong
based on 118
sessions
History
Date
Time
Result
Not Attempted Yet
If p and q are positive integers, what is the remainder when \(9^{2p}*5^{(p+q)} + 11^q*6^{pq}\) is divided by 10?
(A) 0
(B) 1
(C) 3
(D) 4
(E) 5
(A) 0
(B) 1
(C) 3
(D) 4
(E) 5
20 Jan 2022, 08:15
Kudos
Bookmarks
Bunuel
A fast and simple approach is to plug in values of p and q that satisfy the given conditions.
If p and q are positive integers, then it could be the case that p = 1 and q = 1.
Plug these values into the expression to get: \(9^{2p}*5^{(p+q)} + 11^q*6^{pq}=9^{2(1)}*5^{(1+1)} + 11^1*6^{(1)(1)}=9^{2}*5^{2} + 11^1*6^{2}=(81)(25)+(11)(6)=???5 + ???6 = ??????1\)
So, the remainder (when divided by 10) is 1
Answer: B
18 Feb 2022, 07:36
Kudos
Bookmarks
I think it could also be solved using logic - please correct me if I am wrong with my approach!
(9^2p x 5^(p+q)) + (11^q x 6^pq) --> lets look at what happens to the units digits when taking the power of these numbers.
9^2p --> will always have units digit of 9 (consider cyclicity of 9^x - as its 2p the exponent will always be even and thus have a units digit of 1 and not 9)
5^(p+q) --> will always have units digit of 5
Then multiply the units digits of these two terms.
5x9=45 (units digit is 5)
Thus the units digit of the first part (9^2p x 5^(p+q) is certainly 5.
11^q --> will always have a units digit of 1
6^pq --> will always have a units digit of 6
Then multiply the units digits of these two terms.
1x6=6 (units digit 6)
Thus the units digit of the second part (9^2p x 5^(p+q) is 6
Finally add the units digits of both parts together (9^2p x 5^(p+q)) + (11^q x 6^pq)
5+6=11 (units digit 1)
Hence answer is B.
(9^2p x 5^(p+q)) + (11^q x 6^pq) --> lets look at what happens to the units digits when taking the power of these numbers.
9^2p --> will always have units digit of 9 (consider cyclicity of 9^x - as its 2p the exponent will always be even and thus have a units digit of 1 and not 9)
5^(p+q) --> will always have units digit of 5
Then multiply the units digits of these two terms.
5x9=45 (units digit is 5)
Thus the units digit of the first part (9^2p x 5^(p+q) is certainly 5.
11^q --> will always have a units digit of 1
6^pq --> will always have a units digit of 6
Then multiply the units digits of these two terms.
1x6=6 (units digit 6)
Thus the units digit of the second part (9^2p x 5^(p+q) is 6
Finally add the units digits of both parts together (9^2p x 5^(p+q)) + (11^q x 6^pq)
5+6=11 (units digit 1)
Hence answer is B.
