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

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)

=>

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


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)10

This 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)10
Thus b=0

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

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.

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