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# How many positive two-digit integers have a remainder of 1 when divide

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How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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04 Dec 2018, 00:57
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63% (02:27) correct 37% (02:47) wrong based on 103 sessions

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[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$?

$$A. 1$$
$$B. 2$$
$$C. 3$$
$$D. 4$$
$$E. 5$$

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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" NUS School Moderator Joined: 18 Jul 2018 Posts: 1018 Location: India Concentration: Finance, Marketing WE: Engineering (Energy and Utilities) How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags Updated on: 04 Dec 2018, 01:20 1 Let the number be X. The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder. X = 30-1 = 29. General form = 30n-1. where n is integer from 1,2,3... X can have 3 values 29,59 and 89 C is the answer. _________________ Press +1 Kudos If my post helps! Originally posted by Chethan92 on 04 Dec 2018, 01:08. Last edited by Chethan92 on 04 Dec 2018, 01:20, edited 1 time in total. GMAT Club Legend Joined: 18 Aug 2017 Posts: 5289 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags 04 Dec 2018, 01:13 2 1 MathRevolution wrote: [Math Revolution GMAT math practice question] How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$? $$A. 1$$ $$B. 2$$ $$C. 3$$ $$D. 4$$ $$E. 5$$ took >2 mins to solve first list out all 2 digit no which when divided by 5 would give remainder 4 : 14,19,24,29,34,39,44,49,54,59,64,69,74,79,84,89,94,99 all even no as they will be divisible by 2 and no divisible by 3 to be removed : we are then left with 3 no. only : 29,59 & 89 IMO C ... GMAT Club Legend Joined: 18 Aug 2017 Posts: 5289 Location: India Concentration: Sustainability, Marketing GPA: 4 WE: Marketing (Energy and Utilities) Re: How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags 04 Dec 2018, 01:17 Afc0892 wrote: Let the number be X. The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder. X = 30-1 = 29. General form = 30n-1. where n is integer from 1,2,3... X can have 3 values 29,58 and 87 C is the answer. Afc0892 values would be 29,59 & 89 as per formula '30n-1' NUS School Moderator Joined: 18 Jul 2018 Posts: 1018 Location: India Concentration: Finance, Marketing WE: Engineering (Energy and Utilities) Re: How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags 04 Dec 2018, 01:21 Archit3110 wrote: Afc0892 wrote: Let the number be X. The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder. X = 30-1 = 29. General form = 30n-1. where n is integer from 1,2,3... X can have 3 values 29,58 and 87 C is the answer. Afc0892 values would be 29,59 & 89 as per formula '30n-1' Edited. Thank you. _________________ Press +1 Kudos If my post helps! Senior Manager Joined: 04 Aug 2010 Posts: 492 Schools: Dartmouth College Re: How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags 04 Dec 2018, 05:05 1 3 A quick lesson on remainders: Quote: When x is divided by 5, the remainder is 3. In other words, x is 3 more than a multiple of 5: x = 5a + 3. When x is divided by 7, the remainder is 4. In other words, x is 4 more than a multiple of 7: x = 7b + 4. Combined, the statements above imply that when x is divided by 35 -- the LOWEST COMMON MULTIPLE OF 5 AND 7 -- there will be a constant remainder R. Put another way, x is R more than a multiple of 35: x = 35c + R. To determine the value of R: Make a list of values that satisfy the first statement: When x is divided by 5, the remainder is 3. x = 5a + 3 = 3, 8, 13, 18... Make a list of values that satisfy the second statement: When x is divided by 7, the remainder is 4. x = 7b + 4 = 4, 11, 18... The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS: R = 18. Putting it all together: x = 35c + 18. Another example: When x is divided by 3, the remainder is 1. x = 3a + 1 = 1, 4, 7, 10, 13... When x is divided by 11, the remainder is 2. x = 11b + 2 = 2, 13... Thus, when x is divided by 33 -- the LCM of 3 and 11 -- the remainder will be 13 (the smallest value common to both lists). x = 33c + 13 = 13, 46, 79... Onto the problem at hand: Max@Math Revolution wrote: [Math Revolution GMAT math practice question] How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5? A. 1 B. 2 C. 3 D. 4 E. 5 a remainder of 1 when divided by 2 x = 2a + 1 = 1, 3, 5, 7... In other words, x must be ODD. a remainder of 4 when divided by 5 x = 5b + 4 = 4, 9, 14, 19... Since x must be odd, we get: x = 9, 19, 29.... a remainder of 2 when divided by 3 In the blue list above, 29 is smallest value that yields a remainder of 2 when divided by 3. Thus, when x is divided by 30 -- the LCM of the three divisors 2, 3, and 5 -- the remainder will be 29 (the smallest value that satisfies all of the given conditions): x = 30c + 29 = 29, 59, 89, 119... In the resulting list of options for x, only the 3 values in green are two-digit integers. _________________ GMAT and GRE Tutor Over 1800 followers GMATGuruNY@gmail.com New York, NY If you find one of my posts helpful, please take a moment to click on the "Kudos" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. GMATH Teacher Status: GMATH founder Joined: 12 Oct 2010 Posts: 935 Re: How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags 04 Dec 2018, 18:55 MathRevolution wrote: [Math Revolution GMAT math practice question] How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$? $$A. 1$$ $$B. 2$$ $$C. 3$$ $$D. 4$$ $$E. 5$$ $$?\,\,:\,\,\# \,\,N\,\,{\rm{with}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right)$$ $$10\,\, \le \,\,N\,\,{\mathop{\rm int}} \,\, \le \,\,99\,\,\left( 1 \right)$$ $$\left\{ \matrix{ \,N = 2M + 1,\,\,M\,\,{\mathop{\rm int}} \,\,\,\,\left( {2{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( { - 3 \cdot 5} \right)} \,\,\,\,\,\,\, - 15N = - 30M - 15\,\,\,\,\left( {2{\rm{b}}} \right) \hfill \cr \,N = 3K + 2,\,\,K\,\,{\mathop{\rm int}} \,\,\,\,\left( {3{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( {2 \cdot 5} \right)} \,\,\,\,\,\,\,10N = 30K + 20\,\,\,\,\left( {3{\rm{b}}} \right) \hfill \cr \,N = 5L + 4,\,\,L\,\,{\mathop{\rm int}} \,\,\,\,\left( {4{\rm{a}}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\left( {2 \cdot 3} \right)} \,\,\,\,\,\,\,6N = 30L + 24\,\,\,\,\left( {4{\rm{b}}} \right) \hfill \cr} \right.$$ $$\mathop \Rightarrow \limits^{{\rm{sum}}\,\,\left( {{\rm{2b}}{\rm{,3b}}{\rm{,4b}}} \right)} \,\,\,\,N = 30W + 29,\,\,W\,\,{\mathop{\rm int}} \,\,\,\,\,\left\{ \matrix{ \,W = 0\,\,\, \to \,\,\,N = 29\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr \,W = 1\,\,\, \to \,\,\,N = 59\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr \,W = 2\,\,\, \to \,\,\,N = 89\,\,:\,\,{\rm{ok}}\,\,\left( 1 \right),\left( {2{\rm{a}}} \right),\left( {3{\rm{a}}} \right),\left( {4{\rm{a}}} \right) \hfill \cr} \right.$$ $$? = 3$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our high-level "quant" preparation starts here: https://gmath.net Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8152 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: How many positive two-digit integers have a remainder of 1 when divide [#permalink] ### Show Tags 06 Dec 2018, 00:30 => Let $$x$$ be a positive integer with these properties. Since $$x = 2p + 1$$ for some integer $$p$$, the possible values of $$x$$ are $$x = 1, 3, 5, 7, … .$$ Since $$x = 3q + 2$$ some integer $$q$$, the possible values of $$x$$ are $$x = 2, 5, 8, 11, … .$$ Since $$x = 5r + 4$$ for some integer $$4$$, the possible values of $$x$$ are $$x = 4, 9, 14, 19, … .$$ The first possible 2-digit number is thus $$19$$. To find the others, note that the least common multiple of $$2, 3$$ and $$5$$ is $$lcm(2,3,5) = 30.$$ Thus, there are three possible 2-digit numbers with these properties: $$19, 49 = 19 + 30$$ and $$79 = 49 + 30.$$ Therefore, C is the answer. Answer: C _________________ 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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How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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26 Aug 2019, 07:47
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$?

$$A. 1$$
$$B. 2$$
$$C. 3$$
$$D. 4$$
$$E. 5$$

Asked: How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$

N = 1 mod 2 = 2 mod 3 = 4 mod 5
N = 2k1 + 1 = 3k2 + 2 = 5k4 + 4
2k1 + 1 = 3k2 + 2 = odd = {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} = 5 + 6k5
5 + 6k5 = 5k4 + 4 = {29, 59, 89}

IMO C

Alternatively

N = -1 mod 2 = -1 mod 3 = -1 mod 5 = 2k1 -1 = 3k2 -1 = 5k3 -1
= LCM(2,3,5) -1 = 30k -1 = {29,59,89}

IMO C
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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26 Aug 2019, 08:25
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$?

$$A. 1$$
$$B. 2$$
$$C. 3$$
$$D. 4$$
$$E. 5$$

a1= 29

the said series is going to be of the form 30n-1.. so, only 3 numbers possible: 29, 59, 89.
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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23 Oct 2019, 00:05
azhrhasan wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$?

$$A. 1$$
$$B. 2$$
$$C. 3$$
$$D. 4$$
$$E. 5$$

a1= 29

the said series is going to be of the form 30n-1.. so, only 3 numbers possible: 29, 59, 89.

Hi how do you come up with 30n-1 ? I understand that 30 is the LCM but not quite get how to come up with 1 there.
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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23 Oct 2019, 18:32
David nguyen wrote:
azhrhasan wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

How many positive two-digit integers have a remainder of $$1$$ when divided by $$2$$, a remainder of $$2$$ when divided by $$3$$, and a remainder of $$4$$ when divided by $$5$$?

$$A. 1$$
$$B. 2$$
$$C. 3$$
$$D. 4$$
$$E. 5$$

a1= 29

the said series is going to be of the form 30n-1.. so, only 3 numbers possible: 29, 59, 89.

Hi how do you come up with 30n-1 ? I understand that 30 is the LCM but not quite get how to come up with 1 there.

find the first number in the series and express it in terms of 30n + x
Here the first number in series is 29. therefore, 30n-1.

cool ?
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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23 Oct 2019, 18:54
Chethan92 wrote:
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,59 and 89

can you kindly explain why did you take common difference as 1?
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Re: How many positive two-digit integers have a remainder of 1 when divide  [#permalink]

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23 Oct 2019, 20:29
pulak1988 wrote:
Chethan92 wrote:
Let the number be X.
The least possible value of X = LCM(2,3,5)-Common difference between the divisor and the remainder.
X = 30-1 = 29.
General form = 30n-1. where n is integer from 1,2,3...
X can have 3 values
29,59 and 89

can you kindly explain why did you take common difference as 1?

Have a look at the remainders that you get when a number is divided by 2,3 and 5..

A number divided by 2 gives you a remainder of 1.. had the number been one greater, the number would have been completely divisible by 2 leaving no remainder.. the same is the case with 3 and 5.. so in simple words, if the number had been 1 greater, it would be a divisible by 2,3,and 5..

If the number is divisible by 2,3, and 5, then the number should be divisible by 30 .
So the number (N) is of the form-
N + 1 = 30k.. => N = 30K-1
Substitute values of k greater than 0 and you'll get three values of N.. HENCE THE ANSWER IS 3...

hope it helps!

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Re: How many positive two-digit integers have a remainder of 1 when divide   [#permalink] 23 Oct 2019, 20:29
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