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(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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08 Sep 2017, 01:05
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\((7^3)a+(7^2)b+7c+d=1045\), where a, b, c, and d are integers from 0 to 6, inclusively, b=? A. 0 B. 1 C. 2 D. 3 E. 4
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Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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08 Sep 2017, 01:59
If a is 2, the sum will only be 343*2 = 686 Even with b=6, we will only be able to get to 294, the combined sum still around 65 away from 1045. a has to be 3, and 343*3 = 1029 and b has to be zero. c can be 1 or 2, but if c is 1, d will be greater than 6. So c is 2, and d is also 2. Therefore, the value for b is 0(Option A)
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Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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10 Sep 2017, 18:25
=> 1045 = 7*149+2 = 7*(7*21 + 2) +2 = 7*(7*(7*3+0) + 2) + 2 = 7*((7^2)*3 +(7^1)*0 + 2) + 2 = (7^3)*3 +(72)*0 + (7^1)*2 + 2 Thus we have b = 0. Ans: A
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Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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10 Sep 2017, 22:12
MathRevolution wrote: =>
1045 = 7*149+2 = 7*(7*21 + 2) +2 = 7*(7*(7*3+0) + 2) + 2 = 7*((7^2)*3 +(7^1)*0 + 2) + 2 = (7^3)*3 +(72)*0 + (7^1)*2 + 2
Thus we have b = 0.
Ans: A
Can i ask how you even decided to subtract 2? I mean to me it looks just as a guess which can't be a solid solution during gmat instead it can be a great waste of time. Or for all such kind of questions it's a common pattern? Thanks



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Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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23 Sep 2017, 06:04
MathRevolution wrote: \((7^3)a+(7^2)b+7c+d=1045\), where a, b, c, and d are integers from 0 to 6, inclusively, b=? A. 0 B. 1 C. 2 D. 3 E. 4 The answer should be A as follows. \((7^3)a+(7^2)b+7c+d=1045\) can be written as 343a+49b+7c+d=1045. Looking at the above simplified equation we can easily deduce that the first member i.e. 343a is the driving factor in order for the sum to reach 1045 Lets try with a=2. This gives 343*2 = 686. this value is 359 short of the sum. And with maximum permissible value of 6 for the variables can not result into 359 for the rest of the term. Lets try with the value of a=3. This gives 343*3=1029. This is 16 short of the target sum of 1045. No need to look for any value for a>3 as it will increase its contribution by more than 343 which will make the sum far more than 1045. Now we can see unless b=0, 49*b will be minimum 49 which is more than 16 the required short of value. No need to find out corresponding values of c and d. Hence the answer is A (b=0)
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(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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Updated on: 19 Oct 2017, 19:53
cbh wrote: MathRevolution wrote: =>
1045 = 7*149+2 = 7*(7*21 + 2) +2 = 7*(7*(7*3+0) + 2) + 2 = 7*((7^2)*3 +(7^1)*0 + 2) + 2 = (7^3)*3 +(72)*0 + (7^1)*2 + 2
Thus we have b = 0.
Ans: A
Can i ask how you even decided to subtract 2? I mean to me it looks just as a guess which can't be a solid solution during gmat instead it can be a great waste of time. Or for all such kind of questions it's a common pattern? Thanks Hi cbh. below process might help you  Another Approach  \(7^3a+7^2b+7c=1045d\), Now LHS is a multiple of \(7\) so RHS has to be a multiple of \(7\) \(1045\) when divided by \(7\) leaves \(2\) as remainder. Hence \(d=2\). so substituting the value of \(d\) in the given equation we will get \(7^3a+7^2b+7c=1043\). this can be simplified as \(7^2a+7b+c=149\) again \(7^2a+7b=149c\). from the logic explained for \(d\), it is clear that \(c=2\) so we get \(7^2a+7b=1492=147\). The equation can further be reduced to \(7a+b=21\). Now RHS is a multiple of \(7\) so for LHS to be a multiple of \(7\), \(b\) has to be \(0\) Option A
Originally posted by niks18 on 23 Sep 2017, 06:32.
Last edited by niks18 on 19 Oct 2017, 19:53, edited 1 time in total.



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Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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23 Sep 2017, 07:22
MathRevolution wrote: \((7^3)a+(7^2)b+7c+d=1045\), where a, b, c, and d are integers from 0 to 6, inclusively, b=? A. 0 B. 1 C. 2 D. 3 E. 4 If we read this question carefully we see that it is just asking us to convert decimal 1045 or (1045) 10 to a number in base 7 system... (7^3)a+(7^2)b+7c+d=(7^3)a+(7^2)b+(7^1)c+(7^0)d or (abcd) base 7= (1045) base 10 (abcd) 7=(1045) 10This is done by calculating quotients of various powers of 7 involved in 1045; So, 7^3 =343 can get involved maximum 3 times in 1045 what is left is 1045(3*343)=16 7^2=49 can not get involved in 16 7^1=7 can get involved maximum 2 times in 16 what is left is 16(2*7)=02 7^0=1 can get involved maximum 2 times in 02 So..a=3,b=0,c=2 and d=2....and (3022) 7=(1045) 10Thus b=0
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(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
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23 Sep 2017, 10:12
took a while 7^2(7a+b)+7c+d=21*49+16 (as 1045 divided by 49 gives as 21*49 and remainder 16) then by fixing a as 3 (after opening brackets 49*21) and plugging numbers we can after couple rounds come to the conclusion that only when b=0 the equation is satisfied.




(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from &nbs
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