# 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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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.**

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 **

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.