Let's denote the original cheque amount as \( x \) Rupees and \( y \) Paise, where \( x \) and \( y \) are integers ranging from 0 to 99.
The bank teller gave Shailaja \( y \) Rupees and \( x \) Paise instead. Therefore, the amount received was \( 100y + x \) Paise or \( y + \frac{x}{100} \) Rupees.
After buying a toffee for 50 Paise, which is \( \frac{50}{100} = 0.5 \) Rupees, Shailaja was left with \( (y + \frac{x}{100}) - 0.5 \) Rupees.
According to the problem, this remaining amount equals three times the original cheque amount: \( 3(x + \frac{y}{100}) \) Rupees.
We can write the equation:
y+x100-0.5=3(x+y100)
Simplifying further, multiply the whole equation by 100 to eliminate fractions:
100y+x-50=300x+3y
Rearrange terms to isolate variables:
97y=299x+50
Rewriting in terms of integers and solving this linear Diophantine equation:
\( 97y = 299x + 50 \)
We seek \( x \) and \( y \) such that both are less than 100. Testing values near the potential cheque amounts, let's set up test cases for constraints between Rupees 13 and Rupees 14.
Trying \( x = 13 \), the integers from \( 1300 \) to \( 1400 \):
\((97y = 299(13) + 50)\)
\(97y = 3887\)
Solve \( y = \frac{3887}{97} \approx 40.06\), which rounds to \( y = 40 \).
The recovered cheque amount \( x + \frac{y}{100} = 13 + 0.40 = 13.40 \)
The corresponding final cheque statement: 'Over Rupees 13 but less than Rupees 14'.
When $10^{100}$ is divided by 7, the remainder is ?