Find the sum of digits of the least positive integer whose first digit is 7 and which is reduced to a third of its original value when its first digit is transferred to the end.
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The number is 7241379310344827586206896551. The following is not rigorous; it's just some kind of intuition. The idea is like this:
This is repeated until we find a 7 that doesn't have any carry over. (In the fifth step, we will encounter a 7, but without the number itself fully divisible yet (72413 divided by 3 is 24137 with remainder 2), so we cannot stop there.) This gives that number.
There's a better way of solving it. Think about how to set up the algebraic equation, if the number has k digits.
What about 7241379310 ?
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7 2 4 1 3 7 9 3 1 0 / 3 = 2 4 1 3 7 9 3 1 0 3 . 3 3 3 …
We follow a simple algorithm after having made an important conclusion: First, since a number P with 7 at the beginning is thrice of another smaller number s , it must have '24' or '25' or '26' at the beginning because only these three numbers when multiplied by 3 give a number with 7 at the beginning. Now our algorithm is as follows: We start building the number s as well as P from the rightmost place, that is, the unit's place, where we know that there is 7, and at each step we follow the following rule:
A. If the leftmost digit in s is d , and there is some 'carrying digit c above it' (which may be 0) and that in P is d ′ , multiply the leftmost digit d in s with 3 and add c to it.
B. You will get a 2-digit number say 1 0 t + u , where t is the ten's digit and u is the unit's digit. We write u to the left of the d ′ in P , and also write u to the left of d in s .
C. Next we write t as the carrying digit over the leftmost digit u in s , so that we have the transform,
d = u
d ′ = u
c = t
D. Next we return to step A, unless we have got one of the numbers 24, 25 or 26 as the beginning digits of P .
I did this by hand, so here are the first few steps.
s = . . . . 7 and P = . . . . 1
s = . . . . 1 7 and P = . . . . 5 1
s = . . . . 5 1 7 and P = . . . . 5 5 1
.
.
.
.
and so on.
Finally the number P will come out to be 7 2 4 1 3 7 9 3 1 0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 .
P.S. This would make a fine algorithmic question for computer science as well, because brute force search won't work here, due to the sheer magnitude of the number.
Same here exactly.
Use code for solving but logic is the main here
So the answer is 126
I am going to share a tactic that I did and it worked well for this problem, (but you might need a calculator) :
The first digit is 7 and 7/3 = 2.33333 , so the first digit of the number divided by 3 is 2. Thus the first 2 digits are 72. So the first 2 digits are 72. 72/3 = 24. So we know that the first two digit of the number divided by 3 is 24. Thus the first 3 digits of this number are 724.
We repeat this process until it divides perfectly and the last digit is 7. We arrive at 7241379310344827586206896551 after 28 digits of this method.
And the digit sum is 126
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if L is the number of digits of the number then we can write the number as N = 7 ∗ 1 0 L − 1 + x for L > 1 clearly L = 1 (N = 7) is not a solution .... the transformation makes N n e w = x ∗ 1 0 + 7 so 7 ∗ 1 0 L − 1 + x = 3 ∗ ( x ∗ 1 0 + 7 ) 7 ∗ 1 0 L − 1 + x = 3 0 ∗ x + 2 1 x = 2 9 7 ∗ 1 0 L − 1 − 2 1 x ∈ Z + 7 ∗ 1 0 L − 1 ≡ 2 1 % 2 9 ϕ ( 2 9 ) = 2 8 g c d ( 2 9 , 7 ) = 1 1 0 L − 1 ≡ 2 1 ∗ 7 − 1 % 2 9 1 0 L − 1 ≡ 2 1 ∗ 7 ϕ ( 2 9 ) − 1 % 2 9 1 0 L − 1 ≡ 2 1 ∗ 2 5 % 2 9 1 0 L − 1 ≡ 3 % 2 9
the least L which holds the equation true is 28 from which we get x then we get the answer