Home
Class 14
MATHS
2013 xx ? ….. 1 = 62403...

`2013 xx ? ….. 1 = 62403`

A

1

B

2

C

3

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the equation \( 2013 \times ? \ldots 1 = 62403 \), we can follow these steps: ### Step 1: Set up the equation We can express the equation as: \[ 2013 \times x \ldots 1 = 62403 \] where \( x \) is the unknown digit that we need to find. ### Step 2: Isolate \( x \) To find \( x \), we first need to isolate it. We can rewrite the equation as: \[ 2013 \times x = 62403 - 1 \] This simplifies to: \[ 2013 \times x = 62402 \] ### Step 3: Divide both sides by 2013 Now, we will divide both sides by 2013 to solve for \( x \): \[ x = \frac{62402}{2013} \] ### Step 4: Perform the division Now, we perform the division: \[ x = 31 \] ### Step 5: Determine the digit represented by \( ? \) Since \( x \) corresponds to the digit before the last digit (which is 1), we find that the digit represented by \( ? \) is: \[ ? = 3 \] ### Final Answer Thus, the value of \( ? \) is: \[ \boxed{3} \] ---
Promotional Banner

Topper's Solved these Questions

  • MISCELLANEOUS EXERCISE-VIII

    UPKAR PUBLICATION |Exercise QUESTION BANK|81 Videos

  • ODD MAN OUT AND SERIES

    UPKAR PUBLICATION |Exercise QUESTION BANK|75 Videos

Similar Questions

Explore conceptually related problems

What was the day of the week on 15 August 2013? 15 अगस्त 2013 को सप्ताह का कौन सा दिन था?

If a = sqrt((2013)^(2) + 2013 + 2014) , then the value of a is

let A=[[1,1],[0,1]] and P=[[cos(pi/6),sin(pi/6)],[-sin(pi/6),cos(pi/6)]] and Q=PAP^T then P^T Q^2013 P (i) [[1,2013],[0,1]] (ii) [[0,2013],[0,1]] (iii) [[2013,0],[0,2103]] (iv) [[0,2103],[2013,0]]

Capital at the end of year 2012 was Rs. 100 crore and at the end of the year 2013 was Rs. 110 crore. Depreciation during the year 2013 was Rs. 5 crore. Gross investment during 2013 is :

The largest perfect square that divides 2014^(3)-2013^(3)+2012^(3)-2011^(3)+ . . . +2^(3)-1^(3) is

If sum_(n-1)^(2013)"tan"(theta/(2^n))s e c(theta/(2^(n-1)))=tan(theta/(2^a))-tan(theta/(2^b)) then (b+a) equals (a)2014 (b) 2012 (c) 2013 (d) 2014

Find the remainder when 1^(2013)+2^(2013)+3^(2013)+...+2012^(2013) is divisible by 2013

int_(1)^(2013)[(x-1)(x-2)...(x-2013)]dx