Manhatan PS - 600-700

Since the 4 integers on the number line are equally spaced, we can find the distance between integers C and D to find out what equal space each of the 4 integers share with one another.

5^13-5^12= The space between each of the integers A, B, C, and D.

We know that the distance from B to C must also be 5^13-5^12, so:

5^12-B=5^13-5^12
B=2*5^12-5^13

The distance from A to B is also 5^13-5^12, so:

2*5^12-5^13-A=5^13-5^12
A=3*5^12-2*5^13
Factor out a 5^12
5^12[3-2(5)]=-7(5^12) D

It always helps me to take easier examples and see how I would approach it. Sometimes when we see weird numbers etc it throws us off in our thoughts (IMO) so sometimes this idea may work.

2 4 6 8

The difference is 2 ebtween any 2 successive elements(8-6 ,6-4 etc..).

Use the same logic here

A B 5^12 5^13

5^13-5^12 = 5^12(5-1) = 5^12 * 4


From C u can find B wuth this common difference

5^12 - 5^12 (4) = 5^12(1-4) = 5^12 * -3 is B

One more step

5^12 (-3) - 5^12(4) = 5^12 (-3-4) = 5^12 (-7) is A

Regards,
CR

My approach:

We find the distance between C and D: 5^13-5^12
Since we have three of these distances between D and A, we need to subtract these distances from D:
5^13-((5^13-5^12)*3)=> factor 5^12 out=>
5^13-(5^12(5-1)*3)=>5^13-(5^12*12)=> factor 5^12 out
5^12 (5-12)=5^12*(-7)