Question

find the number of three digit numbers if sum of their digits is 10​

Answer 1

Solution :-

Let the first term ' a ' = 109 (Three digit number)

Next term a2 =118

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Last term Tn = 901

d = 118-109 = 9

Now, on solving

Tn = a + ( n - 1 ) d

910 = 109 + (n - 1) 9

910 - 109 = (n - 1)9

801/9= n - 1

89 + 1 = n

n is 90.

Now,

From 100 to 999 the numbers which are excluded (whose sum of digits ≠0) are :-

199, 208,217,226,235,244,253,262,271,280,289,298,307,316,325,334,343,352,361,370,379,388,397,469,478,487,496,559,568,577,586,595,649,658,667,685,694,739,748,757,766,775,784,793,829,838,847,856,865,874,883,892.

So all total 36 numbers are there whose sum of digits ≠ 10.

Therefore, Number of three digit numbers whose sum is 10 = (90-36)= 54.

Answer 2

ANSWER:

• The number of three digit numbers whose sum is 10 = 54.

GIVEN:

• The numbers are all three digit and their sum is 10.

TO FIND:

• The number of three digit numbers whose sum is 10.

EXPLANATION:

100 is the least three digit number. So lets start from 100.

Three digit numbers:

101, 102, 103, 104, 105, 106, 106, 108, 109,.....,999

1 + 0 + 0 = 1

Add 9 to 100 [ As 1 + 9 = 10 ]

100 + 9 = 109

1 + 0 + 9 = 10

The least three digit number whose sum of the digits is 10 is 109.

999 is the highest three digit number.

But after 910 (From 911 upto 999), the sum of the digits will always exceed 10.

So the highest three digit number whose sum of the digits is 10 is 910.

9 + 1 + 0 = 10

$\boxed{\bold{\gray{\large{n = \dfrac{l - a}{d} +1}}}}$

Now here by the formula we can find the number of terms, but we need d(common difference).

$\boxed{\bold{\gray{\large{d = t_2 - t_1}}}}$

$\sf a = t_2 = 109$

Start from 109

109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120

In the given terms after 109, 118 is the least three digit number whose sum of the digits is 10.

1 + 1 + 8 = 10

d = 118 - 109 = 9

a = 109

l = 910

$\sf n = \dfrac{109 - 910}{9} +1$

$\sf n = \dfrac{801}{9} +1$

$\sf n = 89 +1$

n = 90

Lets check our answer.

From 100 to 200:

109, 118, 127, 136, 145, 154, 163, 172, 181, 190, 199

Among these numbers sum of the digits of 199 are not equal to 10.

1 + 9 + 9 = 19 ≠ 10

So 199 is excluded.

From 200 to 300:

208, 217, 226, 235, 244, 253, 262, 271, 280, 289, 298

Among these sum of digits of 289, 298 are not equal to 10.

2 + 8 + 9 = 2 + 9 + 8 = 19 ≠ 10

Now three numbers 199, 289, 298 are excluded.

From 300 to 400:

307, 316, 325, 334, 343, 352, 361, 370, 379, 388, 397.

Now 379, 388, 397 along with 199, 289, 298 are excluded.

In a similar way,

From 400 to 500:

469, 478, 487, 496 are excluded.

From 500 to 600:

559, 568, 577, 586, 595 are excluded.

From 600 to 700:

649, 658, 667, 676, 685, 694 are excluded.

From 700 to 800:

739, 748, 757, 766, 775, 784, 793 are excluded.

From 800 to 900:

829, 838, 847, 856, 865, 874, 883, 892 are excluded.

From 900 to 910 we took only the limited numbers (ie. 901, 910) whose sum of the digits are 10.

The excluded numbers are:

199, 289, 298, 379, 388, 397, 469, 478, 487, 496, 559, 568, 577, 586, 595, 649, 658, 667, 676, 685, 694, 739, 748, 757, 766, 775, 784, 793, 829, 838, 847, 856, 865, 874, 883, 892.

Lets count the excluded numbers.

36 numbers should be subtracted from 90 [ As these are are excluded ]

90 - 36 = 54

HENCE 54 THREE DIGIT NUMBER ARE THERE WHOSE SUM OF THE DIGITS IS EQUAL TO 10.