The sum of the digits of a two number is 8. The number exceeds 17 times the unit’s digit by 2. How do you find the number?

2 Answers
Aug 6, 2016

Answer:

53

Explanation:

Number with two digits can be expressed as :

#10n_(2) + n_(1)# for #n_1, n_2 in ZZ#

We know that the sum of the two digits is 8 so:

#n_1 + n_2 = 8 implies n_2 = 8 - n_1#

The number is 2 more than 17 times the unit digit. We know that the number is expressed as #10n_(2) + n_(1)# while the unit digit will be #n_1#.

#10n_(2) + n_(1) = 17n_1 + 2#

#therefore 10n_2 - 16n_1 = 2#

Substituting:

#10(8-n_1) - 16n_1 = 2#

#80 - 26n_1 = 2#

#26n_1 = 78 implies n_1 = 3#

#n_2 = 8 - n_1 = 8 - 3 = 5#

#therefore# number is #53#

Aug 6, 2016

Answer:

#=53#

Explanation:

Let the unit digit be #y# and ten-digit be #x#
So the number is #10x+y#
So we get
#x+y=8# and
#10x+y=17y+2#
or
#10x+y-17y=2#
or
#10x-16y=2#
Dividing both the sides by 2 we get
#5x-8y=1# From the equation #x+y=8# we get 8x+8y=64
Adding up we get
#5x-8y+8x+8y=64+1#
or
#5xcancel(-8y)+8xcancel(+8y)=65#
or
#13x=65#
or
#x=65/13#
or
#x=5#

By putting the value #x=5# in #x+y=8#
we get
#5+y=8#
or
#y=8-5#
or
#y=3#
Hence the number is #10x+y=10(5)+3=53#