If a is the units digit of 7^47

Question

e-GMAT Question:

If  a  is the units digit of  7^{47}  and b is the rightmost nonzero digit in  (125^{10}× 28^{15}) . What is the value of  a+b ?
A) 1
B) 2
C) 5
D) 6
F) 8

This is  Question 6 of  The e-GMAT Number Properties Marathon

Go to   Question 7 of the Marathon

Hatakekakashi · Accepted Answer

a= unit digit of 7^47
7 has the unit cycle of 4

so 47 mod 4 = 3
and 7^3 = 343 
unit digit 3

a=3

b = right most nonzero digit of 125 ^ 10 x 28 ^ 15

125 = 5^3

125^10 = 5^30

28 = 7 x 4

28^15 = 7^15 x 4^15 
4 = 2^2

28^15 = 7^15 x 2^30

now  (125^{10}× 28^{15})  = 5^30 x 7^15 x 2^30

we know 5x2=10 and the the powers are common so multiply the bases

(125^{10}× 28^{15})  = 10^30 x 7^15

for the right most non digit integer

we need 7^15 unit digit

we know 7 has cycle of 4
15 mod 4 = 3

so 7^15 will also end in 3 ( 7^3 = 343 unit digit 3 )

b=3

a+b= 3+3 
6

(D) imo

pushpitkc · Answer

The cyclicity of number 2 and 7 will be needed to solve the question

-n--------1---2---3---4
-2--------2---4---8---6
-7--------7---9---3---1

Since a is the units digit of  7^{47} , it will be the units digit of  7^{4*11 + 3}  or  7^3 , which is 3.

Similarly, b is the units digit of  (125^{10}× 28^{15})

(125^{10} * 28^{15})  =  (5^3)^{10} * (2^2 * 7)^{15}  =  5^{30}*2^{30}*7^{4*3 + 3}  =  10^{30} * 7^{4*3 + 3}  as  a^m * b^ m = (ab)^m

This will translate to the units digit of  7^3  which is 3.

Therefore, the value of a+b is  6(Option D)

thefibonacci · Answer

7^1 ends in 7
7^2 ends in 9
7^3 ends in 3
7^4 ends in 1
then it repeates
so, 7^47 will have units digit same as 7^3 --> 3 = a

125^10 * 28^15 = 5^30 * 2^30 * 7^14 = 7*14 * 10^30 which will have 30 zeros preceded by units digit of 7^14 (same as that of 7^3 = 3 = b)

hence, a+b=6 (option d).

EgmatQuantExpert · Answer

Solution:

We need to the find the sum of  a  and  b  where,  a  is the units digit of  7^{47}  and  b  is the rightmost nonzero digit in  125^{10}× 28^{15} .
Let’s now calculate  a and  b .

Units digit of 7^{47} 
   47= 4*11+3 
  4*11+3= 4k+3 
 Units digit of  7^(4k+3)  is  3  .
Thus, units digit of  7^{47}  is  3 .
   a= 3  
Rightmost nonzero digit in 125^{10}× 28^{15}
  125^{10}× 28^{15}= (5^3)^10× (2^2*7)^15.
 125^{10}× 28^{15}= 5^30 ×2^30 ×7^15
 125^{10}× 28^{15}= 10^30 ×7^15 
We know, 10^n only gives n number of zeroes at the end of the number. 
Hence, the units digit of 7^15 will be same as the rightmost nonzero digit of 125^{10}× 28^{15}.
   15= 4*3+3 
  4*3+3= 4k+3 
 Units digit of  7^(4k+3)  is  3 . 
Thus, 
   b= 3  
Hence,
   a+b= 3+3 
  a+b=6  
Answer: Option  D

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If a is the units digit of 7^47

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