BTGModeratorVI wrote: ↑

Wed Mar 25, 2020 6:36 am

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

Answer: A
Source: Math Revolution

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

BTGModeratorVI wrote: ↑

Wed Mar 25, 2020 6:36 am

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

Answer: A
Source: Math Revolution

Since n is divisible by 5, the units digit must be 0 or 5. However, since n is also divisible by 4, it must be even, and thus the units digit must be 0. That is, z = 0. Finally, since n is divisible by 3, the sum of the digits of n must be a multiple of 3. Since 3 + 4 + 5 + 0 = 9, which is already a multiple of 3, we need the sum of x and y to be a multiple of 3 also. The largest such number (in our given choices) for n is 345990. However, recall that if a number is divisible by 4, the last two digits of the number has to be divisible by 4 also. So n can’t be 345990 since 90 is not divisible by 4. The next number that fits the criteria (i.e., the last digit of the number is 0 and the sum of the digits is a multiple of 3) is 345960. Since 60 is divisible by 4, we see that the maximum possible value of n is 345960.

Answer: A