If m and n are positive integers and mn is a prime number, which of

Question

If  m  and  n  are positive integers and  mn  is a prime number, which of the following could be the units digit of  4^m + 9^n ?

I. 3
II. 5
III. 7

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

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Bunuel · Accepted Answer

Official Solution:

If  m  and  n  are positive integers and  mn  is a prime number, which of the following could be the units digit of  4^m + 9^n ?

I. 3

II. 5

III. 7

A. I only
B. II only
C. III only
D. I and II only
E. I and III only

" mn  is a prime number, implies that we can have the following four cases:
 
 m=1  and  n=2 ;

m=1  and  n=odd \ prime ;

m=2  and  n=1 ;

m=odd \ prime  and  n=1 .
 
Next, the units digit of 4 in positive integer power follows a repeating pattern of two digits: {4, 6} {4, 6} .... Hence, if  m  is odd, the units digit of  4^m  is 4; if  m  is even, the units digit of  4^m  is 6.

The units digit of 9 in positive integer power also follows a repeating pattern of two digits: {9, 1} {9, 1} .... Hence, if  n  is odd, the units digit of  9^n  is 9; if  n  is even, the units digit of  9^n  is 1.

Thus:
 
If  m=1  and  n=2 , then  4^m + 9^n = 4 + 81 = 85 , making the units digit equal to 5.

If  m=1  and  n=odd \ prime , then  4^m + 9^n = 4 + ...9 = ...13 , making the units digit equal to 3.

If  m=2  and  n=1 , then  4^m + 9^n = 16 + 9 = 25 , making the units digit equal to 5.

If  m=odd \ prime  and  n=1 , then  4^m + 9^n = ...4 + 9 = ...13 , making the units digit equal to 3.
 
Therefore, the units digit of  4^m + 9^n  could be 3 or 5.

Answer: D

Purnank · Answer

Given - m and n are positive integers and mn is a prime number, this will only possible when one of them is 1 and other is prime number, now to check the unit digit of 4^m + 9^n, first lets see the cyclicity for digits 4 and 9.
Cyclicity of both of them is 2
power of 4s =  4 , 1 6 , 6 4 , 25 6 , ... Therefore for odd power unit digit is 4 and for even it is 6. Similarly for 9 odd power is 9 and even power is 1.
Hence if m = 1 and n can be any prime number 2, 3, 5, 7 .....
The possible unit digits of 4^m + 9^n are 3 and 5.
Similarly if n=1 and m can be any prime number 2, 3, 5, 7 ... 
The possible unit digits of 4^m + 9^n are 3 and 5.
Answer D.

Nidzo · Answer

One is told  mn  is a prime number. A prime number is a number which is only divisible by itself and 1, meaning that either  m  or  n  equal 1, while the other is a prime number.

Looking at  4^m + 9^n , one notices that the only unit values that  4^m  can produce are:  4  and  6 , while the only unit values that  9^n  can produce are  9  and  1 .

As either  m  or  n  can equal  1 , there are only 3 possible combinations:

means that there are four possible combinations:
1.  4 + 9 = 13 
2.  4 + 1 = 5 
3.  9 + 9 = 15

From this one sees that the only possible unit's digits are:  3  and  5

ANSWER D

Scholar94 · Answer

Answer: D

from the stem, we can conclude that mn=1×any prime number. This implies that either m or n can be 1 and either m or n can be tye prime number. Also note that we are allowed to add unit digit

Since this is a could be true question, we cab easily plug in number. Let's assume that the prime is 2 or 3. And Let's consider the different scenarios below

Case 1: M=1 and N= 2. Here we can deduce that unit digit of 4^1 is 4 and unit digit of 9^2 is 1. Since we are allow to add unit digit, thus the unit of the expression is 5. Option II could be true.

Case 2: M=3, N=1, 4^3 give unit digit of 4 and 9^1 gives unit digit of 9. 4+9=13. Thus 3 is a possible unit of the expression

Lastly, there is no need to check for the 3rd case again, since we have established that both 1 and 2 could be true and that among option A to E, non of then specify all the 3 cases as possible

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gmatophobia · Answer

As m * n = prime, either m = 1 and n = prime, or n = 1 and m = prime

Let's represent 
 
  m = 4x + p  
  n = 4z + q

Case 1  : m = 1 and n = prime
 
 n = even prime → i.e. n = 2
 
 Unit Digits( 4^m + 9^n ) = 4 + 1 =  5     
 
 
 n = odd prime, → i.e. Remainder(n/4) = 1 or 3
 
 Unit Digits( 4^m + 9^n ) = 4 + 9 = 1 3

Case 2  : n = 1 and m = prime
 
 m = even prime → i.e. m = 2
 
 Unit Digits( 4^m + 9^n ) = 6 + 9 = 1 5     
 
 
 m = odd prime, → i.e. Remainder(m/4) = 1 or 3
 
 Unit Digits( 4^m + 9^n ) = 4 + 9 = 1 3

Hence,   Option D

Sazimordecai · Answer

Your answer is right but I’m just thinking why can’t m be 8 and n be 3 which will be 83 and it’s a prime number . Why must it be that they must either be 1.

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Nidzo · Answer

@ Sazimordecai

You have mistaken  mn  to represent a number where  m  and  n  are digits. Instead,  mn = m*n .  As prime numbers have only two factors, itself and 1, one of  m  or  n  will be a prime number while the other will have to be 1.

Hope this clears things up.

If m and n are positive integers and mn is a prime number, which of

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