Problem
Once A is fixed, the whole parts of A.BC and C.DE are A and A+2, so their sum's whole part is 2A+2.
Reducing five unknown digits to one starting number makes the problem easy to test.
5.NBT.A.1Solve An Easier Related ProblemBecause the five digits are consecutive, the whole-number part of the sum A.BC + C.DE is A + C = 2A + 2, so the single starting digit A controls how big the sum is.
The number written in front of each decimal point is its whole-number part, so A.BC brings A whole units and C.DE brings C whole units.
In a decimal the positions left of the point count whole units while the positions right of it count tenths and hundredths, so the leading digit by itself is the whole amount.
Adding the two decimals adds their whole parts, and since consecutive digits make C = A + 2, that whole total is A + (A + 2) = 2A + 2.
Each decimal breaks cleanly into a whole part and a fractional part, so the two numbers can be added whole-to-whole and fraction-to-fraction and then recombined.
The fractional parts .BC and .DE each sit below the ones place, so together they stay under one whole and never lift the whole-number count above A + C.
Joining A with another A makes two equal groups of A, which is 2A, so A + (A + 2) collects to 2A + 2.
Test A=2: the digits are 2,3,4,5,6, giving A.BC = 2.34 and C.DE = 4.56.
A small guess pinned by the whole-number size is quick to verify.
5.NBT.A.3Guess And Check2.34 + 4.56 = 6.90, which is between 6 and 7, so A=2 is the only fit.
Only one starting digit makes the sum sit in the 6-to-7 window.
5.NBT.B.7Guess And CheckMultiplying by 100 shifts the decimal point two places, turning 2.34 into 234.
Times 100 moves the point two spots, a place-value shift.
5.NBT.B.7Guess And CheckWhen the digits are consecutive, one starting number fixes them all - guess it from the whole-number size, then check!