A 5- digit number has 7 in its tens place. The digit in ones place is 5 less than its digit
at tens place. The hundreds place digit is 4 times the digit at ones place. The digit at
thousands place is the same as hundreds place. Guess the digit at ten thousand place if
the sum of all the digits is 33. Write the number in standard form.
Answers
Answer:
88, 872.
Step-by-step explanation:
Let the 5 digit number be a b c d e.
Each of b,c,d,e are between 0 and 9. a is between 1 and 9.
Given
d = 7. (I solved this question without using this data given).
d - e = 5 ... So e is 0,1,2,3 or 4. d = 5,6,7,8, or 9.
c = e * 4 .... So e can be 0, 1, or 2 at most. c is 0 , 4 or 8.
b = c
a + b + c + d + e = 33
a + 4 e + 4 * e + (e+5) + e = 33
a + 10 e + 5 = 33
a + 10 e = 28
Since a is between 1 and 9, e has to be 2 only.
a = 8 and e = 2.
So c = 8 = b, d = 7.
Number is 88, 872.
:-
88, 872.
Step-by-step explanation:
Let the 5 digit number be a b c d e.
Each of b,c,d,e are between 0 and 9. a is between 1 and 9.
Given
d = 7. (I solved this question without using this data given).
d - e = 5 ... So e is 0,1,2,3 or 4. d = 5,6,7,8, or 9.
c = e * 4 .... So e can be 0, 1, or 2 at most. c is 0 , 4 or 8.
b = c
a + b + c + d + e = 33
a + 4 e + 4 * e + (e+5) + e = 33
a + 10 e + 5 = 33
a + 10 e = 28
Since a is between 1 and 9, e has to be 2 only.
a = 8 and e = 2.
So c = 8 = b, d = 7.
Number is 88, 872.