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Caas
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X is positive interger less than 500.When x is divided by 7 21 Jun 2007, 11:07
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X is positive interger less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
A 18
B 20
C 22
D 24
E 25

Please explain your solution



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X is positive interger less than 500.When x is divided by 7 Updated on: 21 Jun 2007, 12:04
Originally posted by vijay2001 on 21 Jun 2007, 11:30.
Last edited by vijay2001 on 21 Jun 2007, 12:04, edited 1 time in total.
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Answer D 24,

Substituting numbers we get 8, 29, 50, 71.... with a gap of 21 in between

Dividing 500/21 =~23

and first number is 8 so 23+1=24 numbers in total

21 is also LCM of the two numbers 3, 7, may be there is a better way to solve this problem.
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X is positive interger less than 500.When x is divided by 7 Updated on: 21 Jun 2007, 13:30
Originally posted by KillerSquirrel on 21 Jun 2007, 11:33.
Last edited by KillerSquirrel on 21 Jun 2007, 13:30, edited 5 times in total.
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this post was edited !

7*n+1 = x

8,15,22,29 ....

3*n+2 = x

5,8,11,14,17,20,23,26,29 ....

using the AP formula:

progression = 29-8 = 21

491 = 8+(n-1)*21 = 21*n-13

n = 24

the answer is (D)

:-D
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X is positive interger less than 500.When x is divided by 7 21 Jun 2007, 12:05
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Thanks vijay,
But can we somehow avoid substituting numbers?
It's time-consuming.....
Good note about LCM. But I don't know how to use it :(

Squirrel, Please explain why you use AP here. I don't understand your method.
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X is positive interger less than 500.When x is divided by 7 21 Jun 2007, 12:17
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Hey Squirrel, your method seems very nice.
Could you elaborate?
Why do you use AP formula here and how come 491, 497, 478 and 21?
Please explain :) :help2
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X is positive interger less than 500.When x is divided by 7 Updated on: 21 Jun 2007, 13:26
Originally posted by KillerSquirrel on 21 Jun 2007, 12:41.
Last edited by KillerSquirrel on 21 Jun 2007, 13:26, edited 2 times in total.
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hello Caas

I'm sorry to write in such a condensed manner:

using the AP can help you find the number of times a number repeats in an Arithmetic Progression.

since the values for 7*n+1 = x are:

8,15,22,29...

and for

3*n+2 = x are:

5,8,11,14,17,20,23,26,29

then my logic is that 29-8 = 21 is the combined progression.

so we have the first value (8) for the AP formula and the progression (21)

491 = 8+(n-1)*21 = 21*n-13

491 is the greatest possible value for X

:-D
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X is positive interger less than 500.When x is divided by 7 21 Jun 2007, 12:58
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Thanks Squirrel,
I guess we cannot avoid picking numbers here



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