A is a three digit number such that z is the units digit, y is the ten

Question

M is a three digit number such that z is the units digit, y is the tens digit and x is the hundreds digit of M.
Also, M= y*(10z+y), where y and 10z+y are prime numbers. Find x+y+z ?

A. 12
B. 14
C. 17
D. 18
E. 22

Kudos for a correct solution.

nick1816 · Accepted Answer

You are very close to the perfect solution. Though you get the correct answer, there are couple of mistakes in your solution.

1. 9 is not a prime number.
2. We don't have to look for a lot of values.

As y is a prime number, y can be 2,3,5 or 7
When y=2 or 5, 10z+y can never be prime. 
Hence y can be 3 or 7

M=100x+10y+z=y*(10z+y)
 100x+10y+z=10yz+y^2

Hence unit digit of  y^2  is equal to z
z=unit digit of  3^2  or  7^2 
z=9 in both cases

1. when y=3, 10z+y=93, which is not a prime
2. when y=7, 10z+y=97, which is a prime

M= 7*97=679

Peddi · Answer

3 digit number xyz can be written as xyz=100x+10y+z ,given y is prime so the max value of y is 7 ,now plug in values for z which ranges from 1to 9 considering y= 7 ,for z start from the bottom,I am substituting 9, now z= 10*9+7=97,given m=y*(10z+y)= 7 *97= 679 , Sox+y+z= 6+7+9= 22, the answer is E

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jimar · Answer

y is a prime, so possible values of y: 2,3,5,7,9
but (10z+y) is also a prime number number; so possible values of y: 3,7,9 (22,25,32,35 etc can never be prime).
Starting from maximum 2 digit prime number (we can start from min i.e. 13 as well, but we know 13*3 will never be a 3 digit number. we will have to check a lot of values to find first 3 digit number, so we start from back) => 10z+y = 97; 97*7=679 => 6+7+9 = 22

Thus E

Krish728 · Answer

Why can't z be 4?

then We can have 7*47 = 329 = 3+2+9 = 14. What is wrong with this solution?

nick1816 · Answer

If M is 329, value of z is 9. You're getting different values of z, making your solution incorrect.

Krish728 · Answer

I didn't understand. Can you please explain?

nick1816 · Answer

Krish728  Bro you are considering value of z=4

So value of M must be 'xy4', as M='xyz'

But, you are getting value of M= 329(value of z is 9); hence, you are contradicting your initial consideration

iridescent995 · Answer

This is one of the best question I have seen in this forum... though this should have been a 95% difficulty question. 
Thanks for the solution.

KHannigan · Answer

A good solution to get there in 2 minutes might look like this on paper:

M = xyz
M = y(10z + y)

y is prime
(10z + y) is prime
Note - the units digit of 10z + y is y.

From y being a prime digit, we know it must be in {2, 3, 5, 7}

For a multi-digit number to be prime, it's units digit must be in {1, 3, 7, 9}.
So given 10z + y is prime, whose units digit is y, we know that y must be in that set.

The intersection of these two constraints means that y must be in {3, 7}. Two cases to test, but how many permutations of z will we have to test? Hopefully not many.

Ah, another thing: the units digit of a product of two numbers is the product of their units digits. i.e. the units digit of ab * cd == units digit of b*d.
So given that M = y(10z+y), the units digits of M is the units digit of y*y.

Putting this together into a table of possibilities:

y 
 y*y 
 z 
 
  3 
 9 
 9 
 
  7 
 49 
 9

Great, only one option for z.

Testing these possibilities we find:

M = 3 * (10 * 9 + 3) = 279 // Doesn't fit, the y digit comes out as 7 instead of 3.
M = 7 * (10 * 9 + 7) = 679 // This works!

A is a three digit number such that z is the units digit, y is the ten

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A is a three digit number such that z is the units digit, y is the ten

Prep Toolkit

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My Rewards