vinnaymb wrote:Hello All,
Out of touch with maths for 9 years, Sitting somewhere in Saudi (yes...seriously)
need tips..
The question -
what is the sum of all the possible 3 digit numbers that can be constructed using the digits 3, 4 and 5, if each digit can be used only once in each number.
--my answer so far is if I arrange the digits 345,354,435,453,534 and 543..i will get the answer =2664
need tip... how to use shortcut method to get the answer quickly instead of just simply adding these numbers.
In general, if you are looking for shortcuts, you often need to look at the answer choices. I'll give three examples:
345 + 354 + 435 + 453 + 534 + 543 =
A)664
B)1234
C)2664
D)5124
E)9334
then here the answer choices are very far apart. We can then use a rough estimate instead of adding every number. We're adding 6 numbers which are, on average, roughly 450, so the sum should be very close to 6*450 = 2700, and C is the only reasonable answer choice.
345 + 354 + 435 + 453 + 534 + 543 =
A)2579
B)2627
C)2664
D)2685
E)2702
Here the answers are very close together, so estimation is a bad idea. But, the answers all have a different units digit. So if we can find the units digit of the sum, we can pick the right answer. If we add the units digits: 5+4+5+3+4+3 = 24, we can see the answer must end in '4', so C is the only possible answer.
One final possible shortcut, which is more subtle than the others but can be useful:
345 + 354 + 435 + 453 + 534 + 543 =
A)2564
B)2614
C)2664
D)2665
E)2705
Notice we can't easily estimate or use units digits here. But, each number in our sum is divisible by 3 (since the sum of the digits in each number is 3+4+5 = 12, which is divisible by 3). When we add together multiples of 3, we must get a multiple of 3. So we can check each answer choice, by adding the digits, to see which is a multiple of 3. Only answer choice C is a multiple of 3, so it must be correct.
In the particular question you posted, it is not overly time consuming to simply complete the sum, but if you have answer choices available, you can almost always bypass these kinds of tedious computations - but just what technique you might use depends on what the answer choices look like.
