A number between 100 and 200 is 19 times the sum of its digits. The tens digit is 3 more than the units digit. Find the number.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The number given lies between 100 and 200.

It is 19 times the sum of its digits.

Now the multiples of 19 that lie between 100 and 200 are:

114, 133, 152, 171, 190.

We only need to consider these multiples as the sum of the digits is a whole number.

Now use the fact that the tens digit is 3 more than the units digit.

looking at the 5 numbers that we identified earlier only in 152 is the tens digit 3 more than the units digit.

Therefore the required number is 152.

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Given that the number is between 100 and 200,

Then the number is a three digit number.

Then, the hundreds digit is 1.

==> Let the number be  x such that "a" is the tens and "b" is the units.

Then,

x = 100 + 10a + b ..............(1)

Given that 19 times the sum of the digits = x

==>19( 1 + a + b) = x...............(2)

==> 19(1+ a + b) = 100 + 10a + b

==> 19 + 19a + 19b = 100 + 10a +b

==> 9a + 18b = 81

Divide by 9:

==> a + 2b = 9................(3)

Also, given that the tens digit = 3 more that the units.

==> a = 3 + b

We will substitute into (3)

==> (3+b) + 2b = 9

==> 3 + 3b = 9

==> 3b = 6

==> b = 2

==> a = 3+ b = 3+2 = 5

==> a = 5

Then the number is: x = 152

To check:

152 = 19 * (1+5 +2)

152 = 19 * 8

152 = 52

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let the digits of the number between 100 and 200 be 1, x and y.

Then the value of the number = 1*100+10*x+y. The sum of the digits = 1+x+y.

The number is 19 times the sum of the digits. So 100+10x+y = 19(1+x+y)..(1).

The digit in 10 th place is 3 more than the digit in unit place.

 So x= 3+y....(2). So we substitute x= 3+y in (1):

100+10(3+y)+y = 19{1+(3+y)+y}

100+30+10y+y = 19(4+2y) = 76+38y.

130 +11y = 76 +38y.

130-76 = 38-11y = 27y

54 = 27y.

54/27 = 27y/27

2 = y.

So y = 2.

We substitute y= 2 in (2): x = 3+y = 3+2 = 5.

So x= 5, y = 2.

Therefore the number is 152.

Tally : First condition 19 times sum of the digits = number: 19(1+5+2) = 19*8 = 152.

The 2nd condition digit in the 10 th place = 3+the digit in the unit place 5 = 3+2.

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