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How many positive twodigit integers have a remainder of 1 when divide
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04 Dec 2018, 00:57
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[ Math Revolution GMAT math practice question] How many positive twodigit 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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How many positive twodigit integers have a remainder of 1 when divide
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Updated on: 04 Dec 2018, 01:20
Let the number be X. The least possible value of X = LCM(2,3,5)Common difference between the divisor and the remainder. X = 301 = 29. General form = 30n1. where n is integer from 1,2,3... X can have 3 values 29,59 and 89 C is the answer.
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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.



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How many positive twodigit integers have a remainder of 1 when divide
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04 Dec 2018, 01:13
MathRevolution wrote: [ Math Revolution GMAT math practice question] How many positive twodigit 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 ...



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Re: How many positive twodigit integers have a remainder of 1 when divide
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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 = 301 = 29. General form = 30n1. where n is integer from 1,2,3... X can have 3 values 29,58 and 87
C is the answer. Afc0892values would be 29,59 & 89 as per formula '30n1'



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Re: How many positive twodigit integers have a remainder of 1 when divide
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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 = 301 = 29. General form = 30n1. where n is integer from 1,2,3... X can have 3 values 29,58 and 87
C is the answer. Afc0892values would be 29,59 & 89 as per formula '30n1' Edited. Thank you.
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Re: How many positive twodigit integers have a remainder of 1 when divide
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04 Dec 2018, 05:05
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 twodigit 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 2x = 2a + 1 = 1, 3, 5, 7... In other words, x must be ODD. a remainder of 4 when divided by 5x = 5b + 4 = 4, 9, 14, 19... Since x must be odd, we get: x = 9, 19, 29.... a remainder of 2 when divided by 3In 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 twodigit integers.
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Re: How many positive twodigit integers have a remainder of 1 when divide
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04 Dec 2018, 18:55
MathRevolution wrote: [ Math Revolution GMAT math practice question] How many positive twodigit 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.
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Re: How many positive twodigit integers have a remainder of 1 when divide
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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 2digit 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 2digit numbers with these properties: \(19, 49 = 19 + 30\) and \(79 = 49 + 30.\) Therefore, C is the answer. Answer: C
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How many positive twodigit integers have a remainder of 1 when divide
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26 Aug 2019, 07:47
MathRevolution wrote: [ Math Revolution GMAT math practice question] How many positive twodigit 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 twodigit 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 twodigit integers have a remainder of 1 when divide
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26 Aug 2019, 08:25
MathRevolution wrote: [ Math Revolution GMAT math practice question] How many positive twodigit 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 30n1.. so, only 3 numbers possible: 29, 59, 89.



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Re: How many positive twodigit integers have a remainder of 1 when divide
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23 Oct 2019, 00:05
azhrhasan wrote: MathRevolution wrote: [ Math Revolution GMAT math practice question] How many positive twodigit 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 30n1.. so, only 3 numbers possible: 29, 59, 89. Hi how do you come up with 30n1 ? 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 twodigit integers have a remainder of 1 when divide
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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 twodigit 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 30n1.. so, only 3 numbers possible: 29, 59, 89. Hi how do you come up with 30n1 ? 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, 30n1. cool ?



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Re: How many positive twodigit integers have a remainder of 1 when divide
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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 = 301 = 29. General form = 30n1. where n is integer from 1,2,3... X can have 3 values 29,59 and 89
C is the answer. can you kindly explain why did you take common difference as 1?



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Re: How many positive twodigit integers have a remainder of 1 when divide
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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 = 301 = 29. General form = 30n1. where n is integer from 1,2,3... X can have 3 values 29,59 and 89
C is the answer. 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 = 30K1 Substitute values of k greater than 0 and you'll get three values of N.. HENCE THE ANSWER IS 3... hope it helps! Posted from my mobile device




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