X and Y are the digits at the unit's place of the numbers (408X) and

Question

X and Y are the digits at the unit's place of the numbers (408X) and (789Y) where X ≠ Y. However, the digits at the unit's place of the numbers 408X^{63} and
789Y^{85} are the same.What will be the possible value(s) of X+Y ?

A)9

B)10

C)11

D)12

E)None of the Above

A. 9

B. 10

C. 11

D. 12

E. None of the Above

chetan2u · Accepted Answer

This is related to cyclicity..
Now X and Y are different, and unit's digit of  408X^{63}=x^{63}=x^{4*15+3}=x^3  and 
 789Y^{85}=y^{85}=y^{4*21+1}=y^1 
So, x^3 is same as y^1..
Let us check the cyclicity..
Discard, 0,1,5,6 as they are same irrespective of power..
Also Discard 4,9 as when raised to 3 or 1, they are same..
Let us check the remaining..
 2 ....2, 4, 8, 6, so  8  will fit in as 8^1=1, and cyclicity is 8, 4, 2, 6....x+y=2+8=10
 3 .....3, 9, 7, 1 so  7  will fit in as 7^1=1, and cyclicity is 7, 9, 3, 1....x+y=3+7=10

In each case answer is 10

B

nick1816 · Answer

There are two cases possible when 408X^{63} or 408X^{4m+3} have the same unit digit as 789Y^{85} or 789Y^{4n+1}
Case 1 
when X=3 and Y=7 or when X=7 and Y=3
Case 2 
when X=8 and Y= 2 or X=2 and Y=8
 
In both cases X+Y = 10

Sreeragc · Answer

For any difficulties in understanding the solutions in one go, let me make it more clear on why cyclicity of 4 was choosen.
As said in the question the units digit when raised to given powers are same, so we get to know cyclicity is the one in play here.
Now for cyclicities having a repetition of 1 and 2 every numbers are unique and none matches as required in the question. So that's when 
our only consideration concentrates on 4 and the rest of the solution is as provided by the experts above.

Thanks

SirSanguine · Answer

A smarter way of doing this is to find only one possibility after doing the initial few steps ie writing the cyclicity of units place and matching a 4k+1 position with 4m+1..for example (2)^(4k+3)=8^(4k+1), now find X+Y as 10 and mark the answer as we should get the same value for all possible combinations as it is a PS question.

anish777 · Answer

Hello, why did you choose the cyclicity of 4 to begin with?

Bunuel · Answer

Integers ending in 0, 1, 5, and 6 have a cyclicity of 1, meaning that for any positive integer power, the units digit remains the same: 0, 1, 5, and 6, respectively. 
 Integers ending in 2, 3, 7, and 8 have a cyclicity of 4. For 2, the pattern is {2, 4, 8, 6}. For 3, it’s {3, 9, 7, 1}. For 7, it’s {7, 9, 3, 1}. For 8, it’s {8, 4, 2, 6}. 
 Integers ending in 4 or 9 have a cyclicity of 2. If n is odd, a number ending in 4 raised to n will end in 4, and if n is even, it will end in 6: {4, 6}. Similarly, for numbers ending in 9, when n is odd, they end in 9, and when n is even, they end in 1: {9, 1}.

However, notice that even though some numbers have shorter patterns (like cyclicity 2 for 4 and 9), the overall cyclicity remains 4. For example, for numbers ending in 9, the repeating pattern is {9, 1, 9, 1}, which still repeats every 4 terms. So regardless of the specific units digit, we can still apply a cyclicity of 4 when raising them to powers.

X and Y are the digits at the unit's place of the numbers (408X) and

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

X and Y are the digits at the unit's place of the numbers (408X) and

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards