If a two-digit number is added to the number obtained by reversing the order of its digits, then the sum will always by divisible by__________, thus leaving the remainder 0.
please give me varified answer
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If a two-digit number is added to the number obtained by reversing the order of its digits, then the sum will always by divisible by__________, thus leaving the remainder 0.
please give me varified answer
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Step-by-step explanation:
solution of this question is 11
Answer:
Given :-
\large \sf \bullet \: \dfrac{2 - 9z}{17 - 4z} = \dfrac{4}{5} ∙
17−4z
2−9z
=
5
4
Solution :-
On cross multiplication
\large \sf \: 4(17 - 4z) = 5(2 - 9z)4(17−4z)=5(2−9z)
\large \sf \: 68 - 16z = 10 - 45z68−16z=10−45z
\large \sf \: 16z - ( - 45z )= 68 - 1016z−(−45z)=68−10
\large \sf \: 29z = - 5829z=−58
\large \sf \: z \: = \dfrac{ - 58}{29} z=
29
−58
{ \textsf {\textbf{\pink{ \underline{z = - 2}}}}}
z = - 2
Ans. For upper question