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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9022
GMAT 1: 760 Q51 V42
GPA: 3.82
A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 74% (01:25) correct 26% (01:30) wrong based on 82 sessions

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[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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Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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MathRevolution 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

Divisibility by 3:
A & E remaining

Divisibility by 4:
A remaining

A is divisible by 5

IMO A

Posted from my mobile device
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Kinshook Chaturvedi
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Posts: 4887
GMAT 1: 770 Q49 V46
Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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Top Contributor
MathRevolution 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

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
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Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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MathRevolution 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

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
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 9022
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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=>

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 $$345960.$$

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Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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MathRevolution 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

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.

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Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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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. Re: A positive integer n = 345xyz has six digits. What is the maximum poss   [#permalink] 02 Apr 2020, 03:56

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