# Answer Question #4 - E2E Challenge

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### Answer Question #4 - E2E Challenge

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• 1/3

• a = (1/y) is the only real number solution. I believe y=63, so 1/63

• Using the units digit of the decimal equivalent of today’s date in binary as y; which real numbers, a, does the following equation have a unique solution?

a yx + y-x = y

a*3 = 1

so a = 1/3

• a = (1/3)

• a = 1/3

• y=1

a=1

• Posted before completing,

y=1,

equation becomes, (ay-1)x=0

solving, equation will have a unique solution for all real numbers where a IS NOT EQUAL to 1.

• All numbers except 1/3.

• 1/3

• Today’s (yesterday's) date is 11/11/11, which is 00111111b as binary.

This equivalent of 0x3F hexadecimal and 63 decimal.

Therefore, digits of value of 63 are 6 and 3.

So, the finally equation is:

a 63 + 6 - 3 = 6 (or 63a + 6 - 3 = 6)

And it has unique solution:

a = 3/63, which is finally:

a = 1/21

• 0

• a=1/63... ( x(63a-1)=0 since 0b111111 = 63)

• Decimal equivalent of the date "11/11/11" = 63

Units digit of 63 = 3

3x + 3-x = y

Solution: x = 0

• a=1/63