Let the original 2-digit number be xy.
Because 5 times the sum of the digits is 13 less than the number, therefore
5(x + y) = 10x + y - 13
5x + 5y = 10x + y - 13
-5x + 4y = - 13 (1)
The number with reversed digits is yx.
Because 4 times the sum of the digits is 21 less than the reversed 2-digit number, therefore
4(x + y) = 10y + x - 21
4x + 4y = 10y + x - 21
3x - 6y = -21
x - 2y = -7
x = 2y - 7 (2)
Substitute (2) into (1).
-5(2y - 7) + 4y = -13
-10y + 35 + 4y = -13
-6y = -48
y = 8
From (3), obtain
x = 2*8 - 7 = 9
Answers:
The original 2-digit number is 98
The reversed 2-digit number is 89
The difference between the original and the reversed 2-digit numbers is
98 - 89 = 9
Answer:
the answer is B
Step-by-step explanation:
Answer:
Given:-
The algebraic form of an arithmetic sequence is 4n+1.
To find:-
common difference.
first term
remainder when each term of the sequence is divided by 4.
Solution:-
Given,
and n = 1,2,3...
Now,
If n = 2,
The sequence is 5 ,9,13..
Hence, the first term is 5.
Common difference :
=> d =
=> 9- 5
=> d = 4.
Hence, common difference is 4.
Remainder :
=> 5/4 = 4(4) +1
Here, remainder =1
=> 9/4 = 4(2)+1
Here, remainder =1
=> 13/4 = 4 (3)+1
Here , remainder = 1.
Therefore, the remainder when each term of this sequence is divided by 4 is 1 .
Step-by-step explanation: