Understanding the Problem

The question asks for the day of the week of Y's 10th birthday, given that his 7th birthday fell on a Thursday, 10th May, in a leap year. We need to calculate the shift in the day of the week over 3 years.

Calculating the Time Span

The time difference between the 7th birthday and the 10th birthday is:

Number of years = $10 - 7 = 3$ years.

Accounting for Leap Years

The 7th birthday is in a leap year (Year L). The date is 10th May. The 10th birthday is 3 years later, in Year L+3.

The period is from 10th May (Year L) to 10th May (Year L+3). Since 10th May is after February 29th, the leap day of Year L has already passed and does not count within this interval. The subsequent years (L+1, L+2, L+3) are not leap years in the typical 4-year cycle relative to Year L.

Therefore, no leap day (February 29th) occurs between the 7th and 10th birthdays.

Calculating the Day Shift

Each non-leap year adds 1 odd day (365 days = 52 weeks + 1 day).

Total number of days = 3 years $\times$ 365 days/year = 1095 days.

Total number of odd days = Total days $\pmod{7}$

Total odd days = $1095 \pmod{7}$.

Alternatively, since each of the 3 years adds 1 odd day:

Total odd days = $3 \times (365 \pmod{7}) = 3 \times 1 = 3$ odd days.

Determining the Final Day

The day of the week shifts forward by the total number of odd days.

• Starting day (7th birthday): Thursday
• Shift: +3 days
• 10th birthday: Thursday + 3 days = Sunday

Conclusion

Y's 10th birthday will fall on a Sunday.