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

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

Kinshook · Answer

Divisibility by 3: 
A & E remaining

Divisibility by 4: 
A remaining

A is divisible by 5

IMO A

Posted from my mobile device

BrentGMATPrepNow · Answer

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

Archit3110 · Answer

n = 345xyz has six digits and to be divisible by 3,4,5 the no must be factor of 60 or say all three no
starting option A; 345960

for 3 ; sum of no should be divisble by3 ; i.e 27 so YES
for divisibility by 4 ; 345960 should twice divisible by 2; YES
yes for 5 since last digit is 0
IMO A ; is sufficient

MathRevolution · Answer

=>

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

ScottTargetTestPrep · Answer

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

GaurSaini · Answer

Divisibility by 3 : sum of digits is divided by 3
Divisibility by 4 : last two digits are divided by 4
Divisibility by 5 : last digit is either 0 or 5

n = 345xyz

z = either 0 or 5 but odd number can't be divisible by 4, so z = 0.

possible combinations of yz = 00, 20, 40, 60, 80.

Now sum of digits 3+4+5+0 = 12 which is divisible by 3. Hence x+y should be divisible by 3.

Now going through the answer choices, choices B,D,E can be eliminated as last two digits are not divisible by 4. In choice C, x+y = 17 (not divisible by 3). In A, x+y = 15. Hence choice A.

bumpbot · Answer

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

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