Asked by kevi
A two digit numbers is three times the sum of the digits. It is 45 less than the number formed when the digits are interchanged. Find the two digits
Please anyone to help me with solution
Please anyone to help me with solution
Answers
Answered by
bobpursley
let AB be the number. Then
10A+B=3(A+B) or 7A-2B=0
10B+A-45=10A+B or 9A-9B=-45
7A-2B=0
9A-9B=-45 Now, solving that (multipy second by 7 and first by 9)
63A-18B=0
63A-63B=-315
subtract second eq from first.
-18B+63B=315
45B=315
B=7
you solve for A
10A+B=3(A+B) or 7A-2B=0
10B+A-45=10A+B or 9A-9B=-45
7A-2B=0
9A-9B=-45 Now, solving that (multipy second by 7 and first by 9)
63A-18B=0
63A-63B=-315
subtract second eq from first.
-18B+63B=315
45B=315
B=7
you solve for A
Answered by
oobleck
Let the number be written as ab
Then its value is 10a+b
So, now we are told that
10a+b = 3(a+b)
10a+b = 10b+a - 45
Now just crank it out
Then its value is 10a+b
So, now we are told that
10a+b = 3(a+b)
10a+b = 10b+a - 45
Now just crank it out
Answered by
kevi
Please that 10a + b where is it come from in reference to question?
Answered by
oobleck
consider the number 47
The digits are 4,7
its value is 10*4 + 7
better review base-10 numbers
a 3-digit number abc has the value 100a+10b+c
and so on
The digits are 4,7
its value is 10*4 + 7
better review base-10 numbers
a 3-digit number abc has the value 100a+10b+c
and so on
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