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Key concepts:BTGModeratorVI wrote: ↑Wed Mar 25, 2020 6:36 amA 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.BTGModeratorVI wrote: ↑Wed Mar 25, 2020 6:36 amA 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
