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Stamp and Truck Question

1) An envelope has space for at most three stamps. If you are given three stamps of denomination 1, and three stamps of denomination a(a>1) the least positive integer for which there is no stamp value is........

I'm not able to understand what the problem is about...Please help.

2) Ten trucks numbered 1 to 10, are carrying packets of sugar. Each packet weight either 999g or 1000g and each truck carries only packets of equal weights. The combined weight of 1packet selected from the first truck,2 packets selected from the second, 4 packet from the third, and so on and 2^9 packets from the tenth truck is 1022870g. The truck that have the lighter bags are: (A)1,3,5 (B)2,4,5 (C)1,9 (D)2,8

How to do this problem??

Note by Rajath Krishna R
4 years, 2 months ago

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1) The possible values using only 1's are 1, 2, 3 (trivial). The next smallest is \(a\) because after 3 it is not possible to make any higher values without using your denomination \(a\) stamps (make sure you see why). Now the smallest would depend on the value of \(a\). For example, if \(a=5\), then the smallest would be 4 (because 4<a). Similarly, if \(a\ge 5\), then it would still be 4. However, if \(a=4\), then 4 would not work. Also, 5 and 6 would not work because we can still make them with 2 1's and a 4. So 7 would be the least. For \(a=3\), 4 would not work because you can make it with 1 "1" and 1 "3". 4 wouldn't work either. So 5 is the least.

2) If all the trucks had packets of 1000g, then the total weight would be \(1000\times (1+2^1+2^2+\cdots+2^9)=1023000\). Now, since \(1022870<1023000\) we know some of the trucks have packets of 999g. We'll try to choose the right trucks such that subtracting the extra amount of weight we added by assuming they were carrying 1000g will perfectly give us our desired answer. We see that the difference between the two answers are \(1023000-1022870=130\). Therefore we need a sequence of powers of 2 that will add up to be 130. It turns out that \(2^7+2^1=130\). But there is no option with 1,7... where did we go wrong? Note that the amount of packets of the first truck is \(2^0\), NOT \(2^1\). So that means that we have to add one to each of the numbers to fix the off-by-one error. Wee add, and get a final answer of (D) 2,8

Hope I made some sense.

Daniel Liu - 4 years, 2 months ago

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  1. Maybe take the maximum value of all your answers, i.e. 7.

Zi Song Yeoh - 4 years, 2 months ago

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