JEE Main 2025 — 28 January, Evening Shift — Question 21

$$\begin{aligned} &\text{We\; seek\; three-digit\; numbers\; }N=100a+10b+c\; \text{ with\; }a\in\{1,\dots,9\},\;b,c\in\{0,\dots,9\},\\ &\qquad a+b+c=15,\qquad 212\le N\le 999.\\[6pt] &\text{For\; a\; fixed\; }a\text{ let\; }s=15-a,\; \text{ so\; }b+c=s.\\ &\text{The\; number\; of\; nonnegative\; solutions\; }(b,c)\; \text{ with\; }0\le b,c\le9\text{ equals}\\ &\quad \begin{cases} s+1, & 0\le s\le 9,\\[4pt] 19-s, & 10\le s\le 18. \end{cases} \end{aligned}$$

Now list $s=15-a$ for $a=1,\dots,9$:

$$s=14,13,12,11,10,9,8,7,6,$$

so the counts for each $a$ are

$$19-14,\,19-13,\,19-12,\,19-11,\,19-10,\,9,\,9,\,8,\,7 =5,6,7,8,9,10,9,8,7.$$

Hence the total number of three-digit integers with digit sum $15$ is

$$5+6+7+8+9+10+9+8+7=69.$$

We must exclude those $<212$. Such numbers can only have $a=1$ or $a=2$. For $a=1$ there are $5$ numbers (the case $s=14$); for $a=2$ the required $b+c=13$ has no solution with $0\le b\le1$ (since $c$ would exceed $9$). So there are $5$ numbers $<212$ with digit sum $15$.

Therefore the count in the interval $212\le N\le999$ is

$$69-5=64.$$ $$\boxed{64}$$