Question - A set of all 2 digit number with 4 as the sum of the digit.
Plz answer this step by step.
And plz give the correct answer.
I will mark the answer as the brainliest.
Answers
Answer:
Let n=10a+b where a is tens place digit and b is units place digit.
Now, for a
2
+b
3
to be a 2 digit number, the possible combinations of a and b are (1:5,4), (1:8,3), (2:9,2), (3:9,1), (4:9,0) where k:n⟹ all integers from k to n.
For b=4, the units place digit of a
2
+b
3
will always be 4+ units place digit of a non-zero cube whereas units place of n will be 4. Hence, no solution from this set.
For b=3, the units place digit of a
2
+b
3
will always be 7+ units place digit of a non-zero cube whereas units place of n will be 3. Only possibility is a=6⟹n=63 and a
2
+b
3
=63.
For b=2, the units place digit of a
2
+b
3
will always be 8+ units place digit of a non-zero cube whereas units place of n will be 2. Only possibilities are a=2⟹n=22 and a
2
+b
3
=12 or a=6⟹n=62 and a
3
+b
2
=44 which are not feasible.
For b=1, the units place digit of a
2
+b
3
will always be 1+ units place digit of a non-zero cube whereas units place of n will be 1. Hence, no solution exists for this set for this case.
For b=0, the units place digit of a
2
+b
3
will always be 0+ units place digit of a non-zero cube whereas units place of n will be 0. Hence, no solution exists for this set for this case.
Hence, only 1 solution when n=63 exists
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Step-by-step explanation:
because 2 digit number starts from 10
and with sum 4 they will be till 40 only
