SOLUTION: If a = $.01, b = $.02, c = $.03, and so on, what is the value of your first name? Using this alphabet system, one of the days of the week is worth exactly $1.00. Which is it? Expla

Question 1190094: If a = $.01, b = $.02, c = $.03, and so on, what is the value of your first name? Using this alphabet system, one of the days of the week is worth exactly $1.00. Which is it? Explain your reasoning.
-Need help with equation for the second part.

Found 3 solutions by MathLover1, math_tutor2020, ikleyn:
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

A = $.01, B= $.02, C= $.03, D= $.04, E= $.05, F= $.06, G= $.07, H= $.08, I= $.09,
J= $.10, K= $.11, L= $.12, M= $.13, N= $.14, O= $.15, P= $.16, Q= $.17, R= $.18,
S= $.19, T= $.20, U= $.21, V= $.22, W= $.23, X= $.24, Y= $.25, Z= $.26

now, use the letters of your first name, add all numbers of the dollars

one of the days of the week is worth exactly $1.00
Monday->M+o+n+d+a+y=$.13+$.15+$.14+$.04+$.01+$.25=$0.72
Tuesday->$.20+$.21+$.05+ $.19+$.04+$.01+$.25=$0.95
Wednesday->$.23+$.05+$.04+$.14+$.05+$.19+$.04+$.01+$.25=$1.00
Thursday->$.20+$.08+$.21+$.18+$.19+$.04+$.01+$.25=$1.16
Friday->$.06+$.18+$.09+$.04+$.01+$.25=$0.63
Saturday-> $.19+$.01+$.20+$.21+ $.18+$.04+$.01+$.25=$1.09
Sunday ->$.19+$.21+$.14+$.04+$.01+$.25=$0.84
answer: Wednesday

Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

MathLover1 has a great answer, so there's not much I can add in that regard.

Though I can point out a quick way to remove the non-answers.

Let's consider a word like "Monday"
M = 13
O = 15
N = 14
D = 4
A = 1
Y = 25
The numbers refer to the position of each letter in the alphabet, and also the value of the letter in cents

If we focus on just the units digit of each number, then we have
M = 3
O = 5
N = 4
D = 4
A = 1
Y = 5
Then adding said units digits gets us:
3+5+4+4+1+5 = 8+8+6 = 16+6 = 22
The units digit of the answer is what I want to focus on. The '2' means that the sum of the original values will have 2 at the end of the number.
So there's no way to add up to 100 and instead it adds to something of the form xy2 where x and y are digit placeholders for the hundreds and tens place values.

This trick allows us to rule out "monday".

Then Tuesday has these values
T = 20
U = 21
E = 5
S = 19
D = 4
A = 1
Y = 25
Once again we focus on the units digits only
T = 0
U = 1
E = 5
S = 9
D = 4
A = 1
Y = 5
When adding, don't worry about what's going on with the placeholders other than the units digit.
T+U+E+S = 0+1+5+9 = 6+9 = 5 (ignore the tens digit of 1)
D+A+Y = 4+1+5 = 0 (again, only focus on the units digit of the result)
So we have 5+0 = 5 as the units digit of the answer. We can rule out "Tuesday".

The days Thursday through Sunday will have units digits that aren't zero either, allowing us to rule them out. The only thing left is Wednesday which must be the answer (and MathLover1 confirms it as such).

In short, add the units digits and ignore any carry over and the other place values. We're able to remove everything but Wednesday because the final sum has a unit that isn't zero.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

The names of the days all have the common suffix "day".

It costs 4 + 1 + 25 = 30 cents.

So, we should combine 70 cents using the rest of letters.

Since any letter costs not more than 26 cents (and 13.5 cents, in average, because there are 26 letters in English alphabet),
it makes it clear than 3 remaining letters, as in the word "Monday", is not enough.

So, Monday, Friday and Sunday do not work.

Also, thinking this way, it is obvious that the words with 4 or 5 leading letters, like Tuesday or Thursday, have few chances, too.

So, only Wednesday and Saturday are the real candidates to be verified in the first turn,

and it REDUCES the choice to be checked.

It is my semi-joking answer to this joke problem . . .

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