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A positive integer n = 345xyz has six digits. What is the ma

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A positive integer n = 345xyz has six digits. What is the ma

by Max@Math Revolution » Fri Sep 13, 2019 6:56 am
[GMAT math practice question]

A positive integer n = 345xyz has six digits. What is the maximum possible value of n, if n is divisible by 3, 4 and 5?

A. 345960
B. 345970
C. 345980
D. 345985
E. 345990
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by Brent@GMATPrepNow » Fri Sep 13, 2019 7:40 am
Max@Math Revolution wrote:[GMAT math practice question]

A positive integer n = 345xyz has six digits. What is the maximum possible value of n, if n is divisible by 3, 4 and 5?

A. 345960
B. 345970
C. 345980
D. 345985
E. 345990
Key concepts:
A number is divisible by 3 if the sum of its digits is divisible by 3
A number is divisible by 4 if the number created by the last two digits is divisible by 4
A number is divisible by 5 if the units digit is 5 or 0

A quick glance tells us that all 5 numbers are divisible by 5
Now check divisibility by 4

Start with E, the biggest answer choice.
E) Since 90 (the last 2 digits) is not divisible by 4, we know that 345990 is not divisible by 4. ELIMINATE E.
D) Since 85 (the last 2 digits) is not divisible by 4, we know that 345985 is not divisible by 4. ELIMINATE D.
C) 80 (the last 2 digits) IS divisible by 4, so 345980 IS divisible by 4

Let's stay with C.
Is 345980 divisible by 3?
3+4+5+9+8+0 = 29
Since 29 is NOT divisible by 3, we know that 345980 is NOT divisible by 3
ELIMINATE C.

B. 345970
Since 70 (the last 2 digits) is not divisible by 4, we know that 345970 is not divisible by 4
ELIMINATE B.

By the process of elimination, the correct answer is A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Max@Math Revolution » Sun Sep 15, 2019 5:39 pm
=>

If 345xyz is divisible by 3, 3+4+5+x+y+z is divisible by 3, and x+y+z is divisible by 3.
If 345xyz is divisible by 4, the last two digits yz must form a 2-digit number that is divisible by 4. So, 10y+z is divisible by 4.
If 345xyz is divisible by 5, z must be 0 or 5.

In order for 10y + z to be divisible by 4, we must have z = 0. The possible values of y are 0, 2, 4, 6 and 8.
If y=0 and z=0, then the maximum value of x is 9, and 345xyz is 345900.
If y=2 and z=0, then the maximum value of x is 7, and 345xyz is 345720.
If y=4 and z=0, then the maximum value of x is 8, and 345xyz is 345840.
If y=6 and z=0, then the maximum value of x is 9, and 345xyz is 345960.
If y=8 and z=0, then the maximum value of x is 7, and 345xyz is 345780.

The maximum value of 345xyz is 345960.

Therefore, A is the answer.
Answer: A
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