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# help me

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How many 4-digit numbers have the last digit equal to the sum of the first two digits?

Aug 17, 2019

#1
+1

Beginning with 1000, there are 10+9+8+7+6+5+4+3+2+1 = 55 such numbers.
Each additional 1000 [2000, 3000, 4000.....etc] has exactly the same number. Hence:
55 x 9 = 495 four-digit numbers.

Aug 17, 2019
#2
+1

The above answer is accurate for the LAST 3 digits from the RIGHT. I think the question says the LAST digit from the RIGHT = Sum FIRST TWO digits from the LEFT. If that is accurate, then the number would be:
1000s = 90, 2000s = 80, 3000s = 70, 4000s = 60, 5000s = 50, 6000s = 40, 7000s = 30, 8000s = 20, 9000s =10.
So, the total =90+80+70+60+50+40+30+20+10 = 450 such numbers.

Aug 18, 2019
#3
+2

How many 4-digit numbers have the last digit equal to the sum of the first two digits?

The first two can add to 1,2,3,4,5,6,7,8,9

First I am just looking at combinations for the first 2 digits.

01     to     09          9 ways                This line was wrong, there is only 9       10,20,30,40,50,60,70,80,90

11   to      18          8+7 ways                11,12,13,14,15,16,17,18,    21,31,41,51,61,71,81

22    to      27          6+5 ways                22,23,24,25,26,27,             32,42,52,62,72,

33  to     36            4+3 ways                  33,34,35,36,                       43,53,63,

44  to    45              2+1ways                     44,45,                                    54

9+15+11+7+3 = 45

45 combinations for the first two digits

10 combinations for the 3rd digit

The last digit is preset

45*10*1 = 450 possible numbers

OK  Mine is fixed now, i just had a stupid error.     We seem to be in agreement that the answer is 450.

Thanks guest and Rom Aug 18, 2019
edited by Melody  Aug 19, 2019
#4
+1