mental maths

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mental maths

by sumasajja » Fri Jul 29, 2011 8:58 pm
for how many 3 digit whole numbers is the sum of digits equal to 3
a]3
b]5
c]6
d]7
e]8
the answer is
c
any suggestions on how to solve this problem? ,friends!!

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by ntamhane » Fri Jul 29, 2011 9:05 pm
the three digit numbers can be -
300
210
102
111
120

Hence c

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by naveen451 » Sat Jul 30, 2011 7:07 pm
ntamhane wrote:the three digit numbers can be -
300
210
102
111
120

Hence c
and the other number is 201

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by tpr-becky » Sat Jul 30, 2011 7:57 pm
This problem is about finding the pattern - start by writing down the first number you think of - for many people this will be 111 - then think about how that can change - you rearrange the numbers to get a new number but you could use 0 - if you use one 0 then the other two numbers have to be 2 and 1 - methodically write down all the numbers start with numbers that begin with 1 - can only be 102 and 120, then use numbers that start with 2 - can only be 201 and 210. then think if there is any other way and notice that you could also have 2 zeros but the other number would have to be a three - only 300 works (we can't start any numbers with 0 becuase then it wouldn't be a real 3 digit number) - add up all the nubmers to get 6 possible.
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by jaymw » Sat Jul 30, 2011 7:58 pm
for how many 3 digit whole numbers is the sum of digits equal to 3
a]3
b]5
c]6
d]7
e]8
the answer is
c
any suggestions on how to solve this problem? ,friends!!
For such a question it is advisable to start with the biggest or smallest number. Let's start with the smallest number. Since all the numbers have to have 3 digits, the smallest number cannot start with a zero (because then it would be only a 2-digit number).

So, our smallest 3-digit number will begin with a 1. To keep the number as small as possible, the next digit must be zero, and the last digit must bring the checksum up to 3. Hence, that smallest number is 1-0-2.

The next-smallest number cannot start with 1-0-...because for the checksum to be 3, there is only one possible last digit. Thus, we have to increase the second digit as little as possible to get 1-1-1.

By the same logic, our next number has to be 1-2-0. At this point, it is not possible anymore to increase the second digit while keeping the first one as 1. This means that we have to move to the next highest first digit, which is 2.

The smallest 3-digit number starting with a 2 and having a checksum of 3 must be 2-0-1. Once we increase the second digit to 1, we get 2-1-0.

Since we cannot keep the checksum of 3 when further increasing the second digit, we have to increase the first digit to 3. Apparently, the only 3-digit number starting with a 3 and a checksum of 3 is 3-0-0.

Hence, the answer is C.

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by pm » Sat Jul 30, 2011 11:01 pm
Becky Can we use the Seperator Method in this question?