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29 Oct 2009, 14:15
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The integers 1,2,3......,40 are written on the blackboard.The following operation is repeated 39 times. In each repetition, any 2 numbers, say a & b, currently on the blackboard are erased and a new number a+b-1 is written. What will be the number left on the blackboard in the end?
a) 820
b) 821
c) 781
d) 819
e) 780
a) 820
b) 821
c) 781
d) 819
e) 780
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29 Oct 2009, 14:58
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Hi!
I suggest the following approach:
Initially, you should understand that the final number on the blackboard will be "Initial sum of 1-40" + 39*"Result of one operation".
Initial sum of 1-40: \(\frac{(1+n)}{2}*n=\frac{(1+40)}{2}*40=41*20=820\)
Result of one operation:
Erasing two numbers: \(-(a+b)\)
Writing one number: \((a+b)-1\)
As a result, we have \(-(a+b)+(a+b)-1=-1\)
Finally, solve initial equality \(820 + 39*(-1) = 820 - 40 +1 = 781\). So, it should be C.
KUDOS if it were helpful for you
I suggest the following approach:
Initially, you should understand that the final number on the blackboard will be "Initial sum of 1-40" + 39*"Result of one operation".
Initial sum of 1-40: \(\frac{(1+n)}{2}*n=\frac{(1+40)}{2}*40=41*20=820\)
Result of one operation:
Erasing two numbers: \(-(a+b)\)
Writing one number: \((a+b)-1\)
As a result, we have \(-(a+b)+(a+b)-1=-1\)
Finally, solve initial equality \(820 + 39*(-1) = 820 - 40 +1 = 781\). So, it should be C.
KUDOS if it were helpful for you
29 Oct 2009, 15:04
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C) 781.
1+2-1 = 2
2+3-1 = 4
4+4-1 = 7
7+5-1 = 11
11+6-1 = 16
If you added up 1 through 6, you'd get 21. We've done the "operation" 5 times and get 16, which is 5 less than 21. So, add up 1 through 40 and subtract 39 because you've done the operation 39 times which includes subtracting 1.
The sum of numbers in a seris from 1 to n is:
\(\frac{n(n+1)}{2}\) so 40(41)/2 = 20*41 = 820. But we've done the operation which includes subtracting 1 each time, 39 times, so 820 - 39 = 781.
1+2-1 = 2
2+3-1 = 4
4+4-1 = 7
7+5-1 = 11
11+6-1 = 16
If you added up 1 through 6, you'd get 21. We've done the "operation" 5 times and get 16, which is 5 less than 21. So, add up 1 through 40 and subtract 39 because you've done the operation 39 times which includes subtracting 1.
The sum of numbers in a seris from 1 to n is:
\(\frac{n(n+1)}{2}\) so 40(41)/2 = 20*41 = 820. But we've done the operation which includes subtracting 1 each time, 39 times, so 820 - 39 = 781.
papillon86
