must be George and Darwin, since you don't specify whether it is a leap yesr.
But 323-144 = 179 is not a multiple of 7.
So none of the above will always work. Maybe you can check the other gaps to see whether they will work on either a leap year or a non-leap year.
Please help with explanation! Thanks
My four children are called Fiona, Ben, George and Darwin. Their birthdays are on the 15th,35th,144th and 323rd days of the year respectively. Which two of my children have their birthday on the same day of the week?
A Fiona and Darwin
B Ben and George
C George and Darwin
D Darwin and Ben
5 answers
Is option A the right answer?
Considering 1st of Jan as Sunday, then 15th of day of the year falls on Sunday.
So, I just did, 15 + (7×44) = 323
I may be wrong though 😅 I'm curious to know the answer.
Considering 1st of Jan as Sunday, then 15th of day of the year falls on Sunday.
So, I just did, 15 + (7×44) = 323
I may be wrong though 😅 I'm curious to know the answer.
15th day of the year*
Hi Aleks,
Yes , option A is correct.
Could you please explain where (7×44) come from?
Yes , option A is correct.
Could you please explain where (7×44) come from?
omg what- 😭 😂
okay sure!! :D
I took 1st of January as Sunday.
There's a difference of 7 days from one Sunday to the next Sunday and from one Monday to the next Monday and similarly for the rest of the days of the week.
So if 15th day of the year is on Sunday, then what day of the year will be on next Sunday? You just need to add 7 days with 15 to move to the next Sunday.
15 +7 = 22
So the next Sunday will be 22nd day of the year. But none of the sibling' birthday is on "22nd day of the year".
So if you keep on adding 7 with 15, you'll eventually get 323.
That means, I had to add "7", 44 times with 15 to get to 323rd day of the year.
Which also means, exactly 44 weeks after Fiona's birthday, they'll have Darwin's birthday.
If you try this method (adding "7s") with any other siblings' birthdays, like George and Darwin or Fiona and George, it doesn't match. We never get a multiple of 7.
Look at George and Darwin' for instant. George birthday falls on 144th day of the year.
For Darwin' birthday to fall on the same day of the week, there needs to be a multiple of 7. Which never happens here, in this case. 144 + (7x26) = 326 , this crosses Darwin's birthday. Cause Darwin's on 323rd day of the year. So the right option can't be George and Darwin. We never get 323 by adding 7s with 144.
I think that's the same thing what ooooblec meant :))
okay sure!! :D
I took 1st of January as Sunday.
There's a difference of 7 days from one Sunday to the next Sunday and from one Monday to the next Monday and similarly for the rest of the days of the week.
So if 15th day of the year is on Sunday, then what day of the year will be on next Sunday? You just need to add 7 days with 15 to move to the next Sunday.
15 +7 = 22
So the next Sunday will be 22nd day of the year. But none of the sibling' birthday is on "22nd day of the year".
So if you keep on adding 7 with 15, you'll eventually get 323.
That means, I had to add "7", 44 times with 15 to get to 323rd day of the year.
Which also means, exactly 44 weeks after Fiona's birthday, they'll have Darwin's birthday.
If you try this method (adding "7s") with any other siblings' birthdays, like George and Darwin or Fiona and George, it doesn't match. We never get a multiple of 7.
Look at George and Darwin' for instant. George birthday falls on 144th day of the year.
For Darwin' birthday to fall on the same day of the week, there needs to be a multiple of 7. Which never happens here, in this case. 144 + (7x26) = 326 , this crosses Darwin's birthday. Cause Darwin's on 323rd day of the year. So the right option can't be George and Darwin. We never get 323 by adding 7s with 144.
I think that's the same thing what ooooblec meant :))
1 answer