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

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

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13 Sep 2019, 07:18
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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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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" CEO Joined: 03 Jun 2019 Posts: 2935 Location: India GMAT 1: 690 Q50 V34 WE: Engineering (Transportation) Re: A positive integer n = 345xyz has six digits. What is the maximum poss [#permalink] ### Show Tags 13 Sep 2019, 07:37 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 _________________ Kinshook Chaturvedi Email: kinshook.chaturvedi@gmail.com GMAT Club Legend Joined: 11 Sep 2015 Posts: 4887 Location: Canada GMAT 1: 770 Q49 V46 Re: A positive integer n = 345xyz has six digits. What is the maximum poss [#permalink] ### Show Tags 13 Sep 2019, 07:53 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 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 _________________ Test confidently with gmatprepnow.com GMAT Club Legend Joined: 18 Aug 2017 Posts: 6314 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Re: A positive integer n = 345xyz has six digits. What is the maximum poss [#permalink] ### Show Tags 13 Sep 2019, 07:55 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 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] ### Show Tags 15 Sep 2019, 18:07 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: A positive integer n = 345xyz has six digits. What is the maximum poss  [#permalink]

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01 Apr 2020, 14:06
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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02 Apr 2020, 03:56
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