Author 
Message 
TAGS:

Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8141
GPA: 3.82

(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
08 Sep 2017, 01:05
Question Stats:
54% (02:58) correct 46% (02:53) wrong based on 126 sessions
HideShow timer Statistics
\((7^3)a+(7^2)b+7c+d=1045\), where a, b, c, and d are integers from 0 to 6, inclusively, b=? A. 0 B. 1 C. 2 D. 3 E. 4
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 1 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Manager
Joined: 06 Aug 2017
Posts: 78
GMAT 1: 570 Q50 V18 GMAT 2: 610 Q49 V24 GMAT 3: 640 Q48 V29

Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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)
_________________
 Kudos are the only way to tell whether my post is useful.
GMAT PREP 1: Q50 V34  700
Veritas Test 1: Q43 V34  630 Veritas Test 2: Q46 V30  620 Veritas Test 3: Q45 V29  610 Veritas Test 4: Q49 V30  650
GMAT PREP 2: Q50 V34  700
Veritas Test 5: Q47 V33  650 Veritas Test 5: Q46 V33  650



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3308
Location: India
GPA: 3.12

Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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)
_________________
You've got what it takes, but it will take everything you've got



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8141
GPA: 3.82

Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 1 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Retired Moderator
Joined: 25 Feb 2013
Posts: 1164
Location: India
GPA: 3.82

(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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.



Current Student
Joined: 25 Jul 2011
Posts: 57
Location: India
Concentration: Strategy, Operations
GPA: 3.5
WE: Engineering (Energy and Utilities)

Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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
_________________
Please hit kudos button below if you found my post helpful..TIA



Manager
Joined: 24 Jun 2017
Posts: 116

Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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



Manager
Joined: 24 Jun 2017
Posts: 116

(7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
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.



NonHuman User
Joined: 09 Sep 2013
Posts: 13596

Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
Show Tags
19 Oct 2018, 00:12
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: (7^3)a+(7^2)b+7c+d=1045, where a, b, c, and d are integers from
[#permalink]
19 Oct 2018, 00:12






