Math, asked by willzhu12, 19 days ago

On the fourth day of the year, 04/01/2021, the sum of the digits of the day and month was equal to the sum of the digits of the year. How many times will this happen in 2021?

Answers

Answered by INDnaman
9

51 patterns found for Monday, 4 January 2021.

Attachments:
Answered by amitnrw
1

Given : On the fourth day of the year, 04/01/2021, the sum of the digits of the day and month was equal to the sum of the digits of the year.

To Find :  How many times will this happen in 2021

Solution:

04/01/2021

Day 04

Month 01

Sum of Digits of Day and Month  = 0 + 4 + 0 + 1 = 5

Sum of digits of Year = 2 + 0  + 2 +  1  = 5

Sum of digits of Year = 5

Now for Jan    Sum of Digits of month  = 0 + 1 =  1

Sum of Digits of Day should be  5 -  1  = 4

04  , 13  , 22 , 31   ( 4 Days in January)

For Feb  Sum of Digits of month  = 0 + 2=  2

Sum of Digits of Day should be  5 -  2  = 3

03  , 12  , 21   ( 3 Days in February)

For Mar  Sum of Digits of month  = 0 + 3=  3

Sum of Digits of Day should be  5 -  3  = 2

02  , 11  , 20   ( 3 Days in March)

For Apr  Sum of Digits of month  = 0 + 4=  4

Sum of Digits of Day should be  5 -  4  = 1

01  , 10   ( 2 Days in  April)

May , Jun , July , Aug , Sep  has sum of Digit 5 or more hence no possible combination

After That Month of Oct Sum of Digits of month = 1 + 0   = 1

Sum of Digits of Day should be  5 -  1  = 4

04  , 13  , 22 , 31   ( 4 Days in October)

For Nov  Sum of Digits of month  = 1 + 1=  2

Sum of Digits of Day should be  5 -  2  = 3

03  , 12  , 21 , 30    ( 3 Days in November)

For Dec  Sum of Digits of month  = 1 + 2 =  3

Sum of Digits of Day should be  5 -  3  = 2

02  , 11  , 20   ( 3 Days in  Dec)

4 + 3 + 3 + 2  +  4 + 4 +  3  = 23

23  times   this will happen in 2021

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