harpazo Full Member Joined Jan 31, 2013 Messages 891 Jan 29, 2020 #1 The sum of two digits of a 2-digit number is 11. Reversing the digits increase the number by 45. What is the number? Solution: Let a and b = our two digits. a + b = 11 b + a = (a + b) + 45 Is this the correct set up?
The sum of two digits of a 2-digit number is 11. Reversing the digits increase the number by 45. What is the number? Solution: Let a and b = our two digits. a + b = 11 b + a = (a + b) + 45 Is this the correct set up?
firemath Full Member Joined Oct 29, 2019 Messages 602 Jan 29, 2020 #2 Again, the first part is right. But this is the correct second part, similar to Mr. Bland's answer to problem 1: 10x+y=10y+x+45
Again, the first part is right. But this is the correct second part, similar to Mr. Bland's answer to problem 1: 10x+y=10y+x+45
harpazo Full Member Joined Jan 31, 2013 Messages 891 Jan 29, 2020 #3 firemath said: Again, the first part is right. But this is the correct second part, similar to Mr. Bland's answer to problem 1: 10x+y=10y+x+45 Click to expand... I can solve it now but where did 10 come from?
firemath said: Again, the first part is right. But this is the correct second part, similar to Mr. Bland's answer to problem 1: 10x+y=10y+x+45 Click to expand... I can solve it now but where did 10 come from?
firemath Full Member Joined Oct 29, 2019 Messages 602 Jan 29, 2020 #4 If you multiply the 10s digit by 10, then the numeric value ceases to be a place holder and becomes a true value.
If you multiply the 10s digit by 10, then the numeric value ceases to be a place holder and becomes a true value.
harpazo Full Member Joined Jan 31, 2013 Messages 891 Jan 29, 2020 #5 firemath said: If you multiply the 10s digit by 10, then the numeric value ceases to be a place holder and becomes a true value. Click to expand... Ok. Good data.
firemath said: If you multiply the 10s digit by 10, then the numeric value ceases to be a place holder and becomes a true value. Click to expand... Ok. Good data.
