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We define the digit sum of a non-negative integer to it is in the sum of that is digits. For example, the digit amount of (123) is (1+2+3=6).

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How plenty of positive integers less than (100) have digit sum equal to (8)?

It helps to compose all positive integers much less than (100) together two-digit numbers, whereby the an initial digit might be (0).

Let (n) it is in a positive integer v (n.

How many positive integers much less than (100) have actually digit sum equal come (n)?

When (n=1), we have (2) such numbers: (01, 10). As soon as (n=2), we have (3) such numbers: (02, 11, 20). When (n=3), we have (4) together numbers: (03, 12, 21, 30). And so on…

We have the right to see that every time (n) rises by (1), the number of positive integers with digit amount equal come (n) additionally increases by (1).

So there are (n+1) such integers.


Let’s try to answer this concern for (n = 8) again.

What will certainly the end digit of our number with digit-sum (8) be?

If that (0), climate the possibilities are (800, 710, ..., 080). There space (9) together numbers.

If the last digit is (1), climate we have actually (701, 611,..., 071). There are (8) together numbers.

If the final digit is (2), there space (7) such numbers.

And so on… till we have the final digit as (8), once we have (008). Over there is (1) together number.

Thus for (n = 8), there are (0.5 imes 9 imes 10 = 45) such numbers in total.

It is basic to generalise this argument, to display that in the basic case, there space (dfrac(n+1)(n+2)2) together numbers.

How many positive integers between (500) and (999) have actually digit sum equal to (8)?

What can the very first two number be? our table shows us there room ten feasible numbers that can gives us these.


Given the first two digits, we know the last. For this reason the prize is (10).

How plenty of positive integers less than (1000) have digit sum equal come (8), and one number at least (5)?

We deserve to now compose all ours numbers together three-digit numbers, wherein one or both that the first two digits can be (0).

From component (iv), we have actually that there space (10) possibilities in the instance that the hundreds digit is at the very least (5).

By symmetry, there are additionally (10) numbers as soon as the tens digit is at least (5), and another (10) as soon as the units digit is at the very least (5).

These sets space disjoint (have no facets in common), therefore there are (30) such integers altogether.

Alternatively, we can draw up a additional table. We recognize one digit is 5, 6, 7 or 8, and also we know we are including two extra number to this (that might be 0).

The possibilities because that these extra number are very limited, and also it is easy to create down explicitly the feasible numbers.

What is the total of the number sums the the integers native (0) come (999) inclusive?

Each digit from (0) come (9) appears (100) times in the hundreds position.

Each digit from (0) come (9) appears (10) times (10) time in the 10s position.

Each digit from (0) to (9) appears singly (100) time in the units position.

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That renders (300) times altogether because that each digit.

Therefore the full of the number sums is

<300 imes (0+1+2+3+4+5+6+7+8+9)= 13500.>


Oxford University math Aptitude Test, 2013, Q5

Last updated 18-Aug-16
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