Question

Three digit numbers, in which the middle digit is a perfect square, are formed using the digits 1 to 9. Their sum is
A.134055
B.270540
C.170055
D.270504

Answer 1

Hint: Here we have to find the sum of the digits of a number whose middle digits is a perfect square. For that we will find the possible digits that can occur in units, tenth and hundredth place. Then we will find the sum of possible digits in units place, sum of digits in tenth place and sum of digits in hundredth place. We will then add their sum to find the required sum.

Complete step-by-step answer:
It is given that the middle digit of a three digit number is a perfect square.
The digits that can be used in tenth place are 1, 4, 9.
The digits that can be used in units and hundredth place are 1, 2, 3, …….., 9.
Possible choice for tenth place digits \[ = 9 \times 9 = 81\]
Possible choice for units place digits \[ = 3 \times 9 = 27\]
Possible choice for hundredth place digits \[ = 3 \times 9 = 27\]
Sum of digits in tenth place \[ = 81 \times 10 \times \left( {1 + 4 + 9} \right)\]
Adding the terms in the bracket , we get
\[ \Rightarrow \]Sum of digits in tenth place \[ = 81 \times 10 \times 14\]
Multiplying the terms, we get
\[ \Rightarrow \] Sum of digits in tenth place \[ = 11,340\]……... \[\left( 1 \right)\]
Sum of digits in units place \[ = 27 \times 1 \times \left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9} \right)\]
Adding the terms in the bracket, we get
\[ \Rightarrow \] Sum of digits in units place \[ = 27 \times 1 \times 45\]
Multiplying the terms, we get
\[ \Rightarrow \] Sum of digits in units place \[ = 1215\]……… \[\left( 2 \right)\]
Sum of digits in hundredth place \[ = 27 \times 100 \times \left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9} \right)\]
Adding the terms in the bracket, we get
\[ \Rightarrow \]Sum of digits in hundredth place \[ = 27 \times 100 \times 45\]
Multiplying the terms, we get
\[ \Rightarrow \]Sum of digits in hundredth place \[ = 121500\] …………\[\left( 3 \right)\]
Therefore, the sum of all numbers is equal to sum of digits in tenth place, sum of digits in units place and sum of digits in hundredth place.
Therefore,
Sum of all the numbers\[ = 11340 + 1215 + 121500 = 134,055\]
Hence, the correct option is A.

Note: To add the digits 1, 2, 3, ……., we can use the sum formula of an arithmetic series as the given series is forming an arithmetic progression. An arithmetic series is defined as a group of numbers whose difference of the terms from the preceding term is constant. The common difference of the series is 1 here.