basic maths - need tips

[vinnaymb]

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.

[ikaplan]

there is not really a shortcut here, but let's see how you can do it with algebra.

there are six possible numbers that can be created by using 3,4 and 5 only once: 3*2*1=6

in those 6 numbers, digit '3' appears twice (in the place of the hundreds), digit '3' appears also twice in the hundreds, and digit 5 as well; the same principle applies for tens and units.

so you have:
[hundreds]
twice*300= 600+
twice*400= 800+
twice*500=1000
total: 2400

60+
80+
100=
240

6+
8+
10=
24

2400+240+24=2664

[vinnaymb]

Thanks

[Ian Stewart]

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.

[manpsingh87]

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.

digits 3,4,5 can occur at any of the 3 positions i.e. unit position, ten's position and hundred's position.
when 3 is at the unit place - - 3, any of the remaining blank spaces can be filled in 2 ways,
similarly when 4 is at the unit place - - 4, any of the remaining blank spaces can be filled in 2 ways.
also with 5 we have --5 we have 2 ways.
now the sum of the digits at the unit's place would be (3+4+5)*10^0*2=24

similarly when 3 is at the ten's place we have -3- 2 ways of filling up the blank spaces, similarly with 4 and 5 we have 2 ways each.
therefore sum of digits at the ten's place would be (3+4+5)*10^1*2=240

also for filling up hundred's position we have 2 ways each for digits 3,4,5 (3--,4--,5--) therefore sum of digits at the hundred's position is (3+4+5)*10^2*2=2400

therefore required sum=2400+240+24=2664