A logic problem by Christopher Boo
+ 1 □ 1 □ □ 1 □ □ □ 1
Each box contains a non-zero digit, not necessarily distinct from other numbers. Is the equation above possible?
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7 solutions
A great solution. I rather stupidly worked it out by thinking about how the carries would have to work!
It is indeed an useful technique to check the limiting cases both in mathematics and computer science.
This assumes a base 10 number system.
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I would suggest at base 10 number system should be always assumed unless otherwise stated, because problems could have a completely different answer if you could assume any base numbering system. -- Just a thought.
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This problem does not have a different answer, regardless of base chosen. I agree with your suggestion though.
I thought it was numbering system problem.
Argh, did this correctly and accidentally entered the wrong answer. >.<
Fast proof:
9 9 9 < 1 0 0 0
9 9 < 1 0 0
9 < 1 0
0 < 1
so sum of LHS < sum of RHS. And the left-hand side is the maximum value of a sum of a 3-digit number, a 2-digit number, and a 1-digit number.
you did not specify that the numbers were decimal.. in binary the answer is true
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That is clever indeed. But unless otherwise specified, don't we assume digits to be decimal digits?
In binary this is NOT true. Maybe using some 2's complement or odd representation.... but that is not the intent here.
the highest single digit is 9
so, 9 + 9 9 + 9 9 9 = 1 1 0 7 that is less than the summation given(1111).there is 4 short.
so,it is not possible to make 1111 by following the conditions
The cryptogram is not true using binary numbers.... the comment on carrying still applies. In binary using the boxes we would have 1 + 3 + 7 = 11 (base 10) and 1111 = 15 (base 10). 15 -11 = 4..... that is, we are still 4 shy of the sum....
Actually the cryptogram is NOT solvable in any base. Let's call that base N
The sum is 1111(base n) = N 3 + N 2 + N 1 + 1
The boxes can each be at most N P − 1 Where P is the number of boxes in the row.
The first row sums to N-1 The second row sums to N 2 − 1 The third row sums to N 3 − 1 The sum of the three rows is N 3 + N 2 + N − 3 which is 4 less than the addend "1111" at the bottom.
And this process is true for any number of rows M, of boxes arranged in triangular fashion. The addend will always have M+1 digits and be M+1 larger in magnitude than the sum of the boxes in any base N, that you have chosen.
In base 3 for example the boxes add to 2+8+26 = 36 and the "1111" (base 3) is 27+9+3+1 = 40.
This is a nice generalization. Very underrated solution! Thanks for sharing! +1
The last two columns can be filled with any numbers to work. In the first column however, the number (lets call this x) +1 = 11 Since the number cannot be greater than 9, it is impossible.
More exactly, it could be 9 if we were carrying a 2 from the previous row, but the biggest number we can carry from a sum of two digits is 1.
But can't there be a carry which is more than 1?
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Yes for the right hand column you can get a 2 but that doesn't get you anywhere since the largest carry you can get after that is 1 which still leaves you short of 1111.
This seems to me the most obvious and quickest strategy, no need for further addition, unless you really want to!
I thought so also if you us 7 in the right column you can get a sum of 1111
It is a que based on common sense,if we take biggest digits that are all 9s then we will get the sum 1107 which is already smaller than 1111 hence the given sum is already larger than the largest sum,so it proves that we can never achieve that sum(1111)
That was @Anuj Shikarkhane 's solution.
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That is fine. I am sure Rudra came up with it on his own.
Ensure each column adds to 11: 5+5+1 = 11 Carry one 5+5+1 (carry) = 11 The largest number in the final single space can only be 9 so 9+1 (carry) =10 It would only work if the number was 1011
Let's fill the cryptogram with the highest possible values which can be 9 , 9 9 and 9 9 9 .
Their sum is 9 + 9 9 + 9 9 9 = 1 1 0 7 . But the problem says that the sum of the cryptogram is 1 1 1 1 . Therefore, no such configuration is possible which can yield a value of 1 1 1 1 .