The sum of a number of 4 digit having successive digits in increasing order and number formed by

M4maths Previous Puzzles

04 Oct 2010 4 digit number

The sum of a number of 4 digit having successive digits in increasing order and number formed
by taking these 4 digits in reverse order is 12221. Find the number.

OPtion

1) 8765
2) 7654
3) 2345
4) 5432
5) 5678
6) 4567
7) 6789
8) 9876
9) 3456
10) 6543

Solution

{1000x+100(x+1)+10(x+2)+x+3}+{1000(x+3)+100(x+2)+10(x+1)+ x} = 12221
Hence x =4, number is 4567

Correct Option: 6 Best Solution (14)

Dr.Dipin Singh   13 years AGO

If number is ABCD, then D=A+3 and A+D=11 or 2A+3=11 or A=4 .

Then number is 4567

Like? Yes (1) | No   

Pargat Pal Singh   13 years AGO

12221= 4567+ 7654

Like? Yes (1) | No   

Prerna   13 years AGO

From given options , only 7654 and 4567 are such that sum of first and last digit is 11 ( to have at unit's place with 1 as carry).
Out of these 4567 has successive digits in increasing order.

Like? Yes (1) | No   

Arneet   13 years AGO

4567+7654=12221

Like? Yes (1) | No   

S. Singh   13 years AGO

First and last digit should have difference of 3 and sum equal to 11. possible digits are 4 and 7. So number is 4567.

Like? Yes (1) | No   

SUDESHNA SEN   13 years AGO

Let the 4 digits of the number be x, x+1, x+2 and x+3.
Then, {x*1000+(x+1)*100+(x+2)*10+(x+3)} + {(x+3)*1000+(x+2)*100+(x+1)*10+x}=12221
or, 2222x+3333= 12221
or, 2x+3= 11,
or, x= 4.
Therefore, the number is 4*1000+5*100+6*10+7= 4567..... Option 6)

Like? Yes (1) | No   

ASHOK ROY   13 years AGO

Let the 4 digits of the number be x, x+1, x+2 and x+3.
Then, {x*1000+(x+1)*100+(x+2)*10+(x+3)} + {(x+3)*1000+(x+2)*100+(x+1)*10+x}=12221
or, 2222x+3333= 12221
or, 2x+3= 11,
or, x= 4.
Therefore, the number is 4*1000+5*100+6*10+7= 4567..... Option 6)

Like? Yes (1) | No   

YadeeshKumar   13 years AGO

x*1000+(x+1)*100+(x+2)*10+(x+3)=1111x+123
reverse (x+3)*1000+(x+2)*100+(x+1)*10+x=1111x+3210
these two sum
1111x+123+1111x+3210=12221
2222x=8888
x=4
therefore the number is 4567

Like? Yes (1) | No   

JEEVAN DAS   13 years AGO

Let the 4 digits of the number be x, x+1, x+2 and x+3.
Then, {x*1000+(x+1)*100+(x+2)*10+(x+3)} + {(x+3)*1000+(x+2)*100+(x+1)*10+x}=12221
or, 2222x+3333= 12221
or, 2x+3= 11,
or, x= 4.
Therefore, the number is 4*1000+5*100+6*10+7= 4567..... Option 6)

Like? Yes (1) | No   

JAYEETA DESHMUKH   13 years AGO

Let the 4 digits of the number be x, x+1, x+2 and x+3.
Then, {x*1000+(x+1)*100+(x+2)*10+(x+3)} + {(x+3)*1000+(x+2)*100+(x+1)*10+x}=12221
or, 2222x+3333= 12221
or, 2x+3= 11,
or, x= 4.
Therefore, the number is 4*1000+5*100+6*10+7= 4567..... Option 6)

Like? Yes (1) | No   

Brij   13 years AGO

Suppose the number is XXXX+0123 , then
XXXX+0123+XXXX+3210 = 12210 ; 2(XXXX) = 8888; X =4444 and number is 4444+0123=4567

Like? Yes (1) | No   

midhun prakash   13 years AGO

4567+7654(reverse) =12221

Like? Yes (1) | No   

Garima   13 years AGO

4567+7654=12221

Like? Yes (1) | No   

Achelal Sharma   13 years AGO

THE NUMBER IS 4567 + 7654 = 12221

Like? Yes (1) | No